Quadrilaterals - Class 9 Mathematics - Chapter 8 - Notes, NCERT Solutions & Extra Questions
Renews every month. Cancel anytime
Your personal doubt-solving assistant
Chatterbot AI gives you 100% accurate answers to your questions in an instant.
Extra Questions - Quadrilaterals | NCERT | Mathematics | Class 9
A roller is in the shape of:
Option A) Cuboid
Option B) Cone
Option C) Cylinder
Option D) Sphere
The correct answer is:
Option C) Cylinder
A roller is typically in the shape of a cylinder, which has two circular bases connected by a curved surface, allowing it to roll smoothly along a flat surface.
In the below figure, if $AP = BP$ and $AQ = BQ$, then is $AM = BM$?
A) True
B) False
C) Data insufficient
D) Cannot be determined
In the provided figure, both $AP = BP$ and $AQ = BQ$ are given. Additionally, line segment $PQ$ is common to both triangles $\triangle PAQ$ and $\triangle PBQ$. By applying the Side-Side-Side (SSS) congruence theorem, we can establish that:
$$ \triangle PAQ \cong \triangle PBQ $$
Consequently, by Corresponding Parts of Congruent Triangles are Congruent (CPCT), it follows that:
$$ \angle APQ = \angle BPQ $$
Furthermore, given that $AP = BP$ and considering the shared line segment $PM$, we can apply the Side-Angle-Side (SAS) congruence rule to deduce that:
$$ \triangle PMA \cong \triangle PMB $$
Thus, from the congruence of these triangles, it directly follows that: $$ AM = BM $$ Therefore, the correct answer is A) True.
The fourth vertex $D$ of a parallelogram $ABCD$ whose three vertices are $A(-2,3)$, $B(6,7)$, and $C(8,3)$ is:
(A) $(0,1)$
(B) $(0,-1)$
(C) $(-1,0)$
(D) $(1,0)$
To determine the fourth vertex $D$ of a parallelogram $ABCD$ where the vertices $A$, $B$, and $C$ are given as $A(-2,3)$, $B(6,7)$, and $C(8,3)$, we can utilize the property that diagonals of a parallelogram bisect each other.
Let $D$ have coordinates $(x_4, y_4)$. The midpoints of diagonals $AC$ and $BD$ should be identical given the parallelogram’s properties.
Calculating the midpoint $L$ of diagonal $AC$: $$ L = \left(\frac{-2+8}{2}, \frac{3+3}{2}\right) = (3, 3) $$ Now, let's calculate the midpoint $M$ of diagonal $BD$, where $D$ is $(x_4, y_4)$: $$ M = \left(\frac{6+x_4}{2}, \frac{7+y_4}{2}\right) $$ Setting $L$ equal to $M$ because diagonals bisect each other: $$ 3 = \frac{6 + x_4}{2} \quad \text{and} \quad 3 = \frac{7 + y_4}{2} $$ Solving these equations, we find: $$ 6 = 6 + x_4 \quad \Rightarrow \quad x_4 = 0 $$ $$ 6 = 7 + y_4 \quad \Rightarrow \quad y_4 = -1 $$ Thus, the coordinates of $D$ are $(x_4, y_4) = (0, -1)$. Therefore, the fourth vertex of the parallelogram $ABCD$ is: $$ D = (0, -1)$$ This corresponds to option (B) $(0, -1)$.
In the parallelogram $ABCD$, $AB=6$ cm and $DC:CB=2:1$. Find the perimeter of the parallelogram $ABCD$.
A) $17$ cm
B) $18$ cm
C) $19$ cm
D) $20$ cm
In a parallelogram, opposite sides are equal and parallel.
Given:
$AB = 6$ cm
$DC:CB = 2:1$
From the properties of a parallelogram:
$$ AB = CD = 6 , \text{cm} $$
Since $DC:CB = 2:1$, we can write:
$$ \frac{DC}{CB} = \frac{2}{1} \ \Rightarrow \frac{6}{CB} = \frac{2}{1} \ \Rightarrow CB = \frac{6 \times 1}{2} = 3 , \text{cm} $$
Also, in the parallelogram $ABCD$:
$$ AD = BC = 3 , \text{cm} $$
The perimeter of parallelogram $ABCD$ is calculated as:
$$ \text{Perimeter} = AB + BC + CD + DA = 6 , \text{cm} + 3 , \text{cm} + 6 , \text{cm} + 3 , \text{cm} = 18 , \text{cm} $$
The correct answer is $\textbf{B) 18 cm}$.
If $PQRS$ is a parallelogram, with point $A$ on $QR$, then $\frac{AP}{AB}$ is equal to
A) 1
B) 2
C) $\frac{1}{2}$
D) Can't be determined
The correct choice is Option A: 1.
Observing parallelogram $PQRS$, we know opposite sides are parallel. This makes $PS$ parallel to $RQ$ and $PS$ also parallel to $RA$ (since $A$ is on $RQ$).
Looking at $\triangle BPS$ and applying the Basic Proportionality Theorem (also known as the Thales theorem in some contexts), we establish a proportion related to the segments. Since line $RA$ is parallel to $SP$ and transversal line $AB$ intersects them, we can set up the following proportion based on segment ratios: $$ \frac{BR}{RS} = \frac{AB}{AP} $$ Given that $RS = x$ (where $x$ is the length $RS$) and $BR = x$ (since $PQRS$ is a parallelogram and $PR = BQ$ implies $BR = RS$), the equation becomes: $$ \frac{x}{x} = \frac{AB}{AP} $$ This simplifies to: $$ 1 = \frac{AB}{AP} $$ Thus, $$ \frac{AP}{AB} = 1 $$ Therefore, $\frac{AP}{AB} = 1$ aligns with Option A.
The number of diagonals in an octagon is:
A) 28
B) 14
C) 16
D) 20
To determine the number of diagonals in an octagon, we use the general formula for the number of diagonals in a polygon: $$ \text{Number of diagonals} = \frac{n(n-3)}{2} $$ where $n$ is the number of sides of the polygon. For an octagon, $n = 8$. Substituting into the formula gives: $$ \text{Number of diagonals} = \frac{8(8-3)}{2} = \frac{8 \times 5}{2} = 20 $$ Thus, the correct answer is Option D: 20.
In the given figure, $PQRS$ is a parallelogram and $PQ=8$ cm. If the area of triangle $PXQ$ is $32$ sq cm, find the altitude of the parallelogram.
A) $10$ cm
B) $4$ cm
C) $8$ cm
D) $12$ cm
The correct option is C) 8 cm.
Given the area of triangle $PXQ$, which is half of the parallelogram $PQRS$ since both share the same base $PQ$ and are between the same parallels, we set up the equation: $$ \text{Area of } \triangle PXQ = \frac{1}{2} \times \text{Area of parallelogram } PQRS $$ Where,
Area of $\triangle PXQ = 32 \text{ cm}^2$
$PQ = 8 \text{ cm}$
From the area of the triangle, the formula is: $$ 32 = \frac{1}{2} \times PQ \times h $$
Solving for the altitude $h$: $$ 64 = 8 \times h $$ $$ \frac{64}{8} = h $$ $$ h = 8 \text{ cm} $$
Therefore, the altitude of the parallelogram is 8 cm.
A quadrilateral is drawn such that its diagonals intersect each other in the interior. What type of quadrilateral is this?
A) Concave
B) Convex
C) Closed figure
D) None of the above
The correct answer is B) Convex.
A convex quadrilateral is defined as a quadrilateral where the line segment joining any two non-adjacent vertices lies entirely inside the quadrilateral. This characteristic ensures that when diagonals are drawn in a convex quadrilateral, they intersect each other within the interior of the quadrilateral.
For instance, in the case of a parallelogram $ABCD$, the diagonals intersect each other inside the quadrilateral, demonstrating the aforementioned property typical of convex quadrilaterals.
Fill in the blank to make the statement true.
If the sum of the number of vertices and faces in a polyhedron is 14, then the number of edges in that shape is $\qquad$
Given the sum of the number of vertices ($V$) and faces ($F$) in a polyhedron is 14: $$ V + F = 14 $$
We utilize Euler's formula for polyhedra, which states: $$ F + V - E = 2 $$
Substituting the given sum into Euler's formula: $$ 14 - E = 2 $$
Solving for the number of edges ($E$): $$ E = 14 - 2 \ E = 12 $$
Therefore, the number of edges in that shape is 12.
Construct a cyclic quadrilateral $ABCD$ when $AB = 8$ cm, $BC = 6$ cm, $\angle ABC = 60^{\circ}$, and $AD = 5.4$ cm.
To construct the cyclic quadrilateral $ABCD$ with given dimensions $AB=8$ cm, $BC=6$ cm, $\angle ABC=60^\circ$, and $AD=5.4$ cm, follow the steps below:
Construction Steps
Sketch the Base: Draw a line segment $AB=8$ cm.
Angle Creation: At point $B$, construct an angle $\angle ABC=60^\circ$. Extend a line from $B$ to point $X$ accordingly.
Arc for Side BC: Using $B$ as the center and a radius of 6 cm, draw an arc to cut the line $BX$ at point $C$. Connect $AC$.
Construct the Circumcenter: Draw the perpendicular bisectors of $AB$ and $BC$. Let these bisectors intersect at point $O$, which becomes the circumcenter of $\triangle ABC$.
Draw the Circumcircle: With $O$ as the center and radius $OA$ (which equals $OB = OC$), draw the circumcircle passing through $A$, $B$, and $C$.
Locate Point D: From $A$ as the center, using a radius of 5.4 cm, draw another arc to intersect the previously drawn circumcircle at point $D$.
Complete the Quadrilateral: Lastly, connect $AD$ and $CD$.
Following these steps, $ABCD$ is the desired cyclic quadrilateral, correctly incorporating all the specified measurements and angle.
The length of the longest bar that can be fixed in a room of dimensions $12 , \mathrm{m} \times 9 , \mathrm{m} \times 8 , \mathrm{m}$ is $\quad$ m.
Consider applying a three-dimensional perspective on the situation. Imagine you are inside a room shaped like a cuboid with dimensions as given, namely length $12 , \mathrm{m}$, width $9 , \mathrm{m}$, and height $8 , \mathrm{m}$.
Firstly, examine a vertical face of the cuboid - let's select the face formed by the length and the width. Label the corners of this rectangular face as $A$, $B$, $C$, and $D$. Here, assume $AB = 9 , m$ and $BC = 12 , m$. Since angle $B$ is a right angle, triangle $ABC$ forms a right triangle. The diagonal $AC$ of this face, which is the hypotenuse of $\triangle ABC$, can be computed using the Pythagorean theorem:
$$ \begin{align*} AC^2 &= AB^2 + BC^2 \ &= 9^2 + 12^2 \ &= 81 + 144 \ &= 225 \ AC &= \sqrt{225} \ &= 15 , \mathrm{m} \end{align*} $$
Moving now to a three-dimensional consideration, consider the longest bar that can stretch from one corner to the exactly opposite corner of the cuboid. Picture this as a diagonal extending from corner $A$, intersecting at $C$, and reaching to a corner $E$ directly above $C$ at ceiling level.
From the base diagonal $AC$, extend vertically to the ceiling. Triangle $ACE$ (with $A$ at one base corner, $E$ at the opposite top corner, and $C$ at the other base corner directly below $E$) forms a right triangle where $AE$ is the hypotenuse. By using the Pythagorean theorem applied to $\triangle ACE$:
$$ \begin{align*} AE^2 &= AC^2 + CE^2 \ &= 15^2 + 8^2 \ &= 225 + 64 \ &= 289 \ AE &= \sqrt{289} \ &= 17 , \mathrm{m} \end{align*} $$
Thus, the longest bar that can be fitted inside the room measures $17 , \mathrm{m}$.
Find the side of the square whose perimeter is $20$ m.
The perimeter of a square is defined as the total length around the square. For a square, the perimeter is given by:
$$ \text{Perimeter} = 4 \times \text{side length} $$
Given the perimeter of the square is $20$ m, we can determine the side length by setting up the equation:
$$ 20 = 4 \times \text{side length} $$
Solving for the side length, we divide both sides by $4$:
$$ \text{side length} = \frac{20}{4} = 5 \text{ m} $$
Therefore, the length of each side of the square is 5 m.
The area of a triangle with vertices $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $C\left(x_{3}, y_{3}\right)$ is:
A) $\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$
B) $\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{3}\left(y_{3}-y_{1}\right)+x_{1}\left(y_{1}-y_{2}\right)\right]$
C) $\frac{1}{2}\left[x_{2}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$
D) All of the above.
Solution:
The correct answer is A:
$$ \frac{1}{2}\left[x_1\left(y_2-y_3\right) + x_2\left(y_3-y_1\right) + x_3\left(y_1-y_2\right)\right] $$
Consider a triangle $ABC$ where the vertices are $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. From each vertex, draw perpendiculars to the $x$-axis, denoted as $AP$, $BQ$, and $CR$. Observe the resulting trapeziums formed — $ABQP$, $APRC$, and $BQRC$.
The area of triangle $ABC$ can be disaggregated into these geometric figures:
Area of $ABC$ = Area of trapezium $ABQP$ + Area of trapezium $APRC$ - Area of trapezium $BQRC$.
The area of a trapezium can be calculated as: $$ \text{Area} = \frac{1}{2} (\text{Sum of parallel sides}) \times (\text{Distance between the parallel sides}) $$
Applying this to our trapeziums and consolidating, we get:
$$ \begin{aligned} \text{Area of } \triangle ABC &= \frac{1}{2}(BQ + AP)QP + \frac{1}{2}(AP + CR)PR - \frac{1}{2}(BQ + CR)QR \ &= \frac{1}{2}\left(y_2 + y_1\right)\left(x_1 - x_2\right) + \frac{1}{2}\left(y_3 + y_1\right)\left(x_3 - x_1\right) - \frac{1}{2}\left(y_2 + y_3\right)\left(x_3 - x_2\right) \ &= \frac{1}{2}\left[x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right)\right] \end{aligned} $$
Thus, the area of $\triangle ABC$ is correctly expressed by:
$$ \frac{1}{2}\left[x_1\left(y_2 - y_3\right) + x_2\left(y_3 - y_1\right) + x_3\left(y_1 - y_2\right)\right] $$
Conclusion: The formula stated in Option A is the correct expression for calculating the area of a triangle with given vertices in a coordinate plane.
In the figure, $ABCD$ is a parallelogram in which $\angle ADC=105^\circ$. The value of $\angle CBE$ is
A) $105^\circ
B) $75^\circ
C) $55^\circ
D) $90^\circ
The correct answer is Option B, which is $ 75^\circ $.
Given that $ABCD$ is a parallelogram, it follows that opposite angles are equal. Therefore, $$ \angle ADC = \angle ABC = 105^\circ. $$
Since $$ \angle ABC + \angle CBE = 180^\circ $$ (as they form a linear pair), we can compute $\angle CBE$ as follows: $$ \begin{aligned} \angle CBE &= 180^\circ - \angle ABC \ &= 180^\circ - 105^\circ \ &= 75^\circ. \end{aligned} $$
Thus, $\angle CBE = \textbf{75}^\circ$ is the final answer.
In which of the following quadrants does the point $(-6,1)$ lie?
A) I quadrant.
B) II quadrant.
C) III quadrant.
D) IV quadrant.
The correct option is B) II quadrant.
The coordinates given are $(-6,1)$. Here, the abscissa (x-coordinate) is negative and the ordinate (y-coordinate) is positive. This indicates that the point is located to the left of the y-axis (due to the negative x-coordinate) and above the x-axis (due to the positive y-coordinate). Consequently, this point is positioned in the II quadrant, where all points have negative x-coordinates and positive y-coordinates.
Steps for the construction of a quadrilateral $PQRS$ in which $PQ=4$ cm, $QR=3$ cm, $PS=2.5$ cm, $PR=4.5$ cm, and $QS=4$ cm are given below.
Choose the correct order:
- Draw a line segment $PQ=4$ cm.
- With 4.5 cm from $P$, draw an arc.
- Take 2.5 cm from $P$ and draw an arc.
- From $Q$, take radius 3 cm and cut the arc drawn in step 2. Let the point of intersection be $R$.
- Take 4 cm from $Q$ and cut the arc drawn in step 4. Let the point of intersection of arcs be $S$. Join $SR$.
(A) $1, 2, 4, 3, 5$ (B) $1, 2, 3, 4, 5$ (C) $1, 2, 5, 4, 3$ (D) $1, 2, 3, 5, 4$
The correct sequence to construct quadrilateral $PQRS$ is Option A: 1, 2, 4, 3, 5. Below, I'll describe each step in detail:
-
Step 1: Draw a line segment $PQ = 4$ cm.
-
Step 2: From point $P$, use a compass set to 4.5 cm to draw an arc.
-
Step 4: From point $Q$, with your compass set to 3 cm, draw an arc to intersect the arc from Step 2. Label the intersection as point $R$.
-
Step 3: Now, go back to point $P$. Set your compass to 2.5 cm and draw an arc.
-
Step 5: Adjust the compass to 4 cm and from point $Q$, draw an arc to intersect the arc from Step 4. Label this intersection as point $S$. Finally, connect points $S$ and $R$.
This order of steps ensures all sides of the quadrilateral meet their specified lengths, thereby enabling accurate construction of quadrilateral $PQRS$.
The diagonals of a rhombus are $18 \mathrm{~cm}$ and $24 \mathrm{~cm}$. Find: (i) its area (ii) length of its sides (iii) its perimeter
Solution
Given diagonals of the rhombus are $18 , \text{cm}$ and $24 , \text{cm}$.
(i) Area of the rhombus can be computed using the formula: $$ \text{Area} = \frac{1}{2} \times \text{(Product of diagonals)} = \frac{1}{2} \times 18 \times 24 = 216 , \text{cm}^2 $$
(ii) Since the diagonals of a rhombus bisect each other at right angles: $$ OA = \frac{1}{2} \times 24 = 12 , \text{cm} \ OB = \frac{1}{2} \times 18 = 9 , \text{cm} $$
In $\triangle OAB$, which is a right triangle: $$ AB = \sqrt{OA^2 + OB^2} = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 , \text{cm} $$ Length of each side of the rhombus is therefore $15 , \text{cm}$.
(iii) Perimeter of the rhombus can be calculated as: $$ \text{Perimeter} = 4 \times \text{side} = 4 \times 15 = 60 , \text{cm} $$
Results:
- (i) Area: $216 , \text{cm}^2$
- (ii) Side Length: $15 , \text{cm}$
- (iii) Perimeter: $60 , \text{cm}$
Mention any four properties of a square. [4 MARKS]
Solution:
Each property of a square detailed below earns 1 Mark:
a) Opposite sides are parallel: This geometric characteristic ensures that each pair of facing sides in a square runs exactly in the same direction, never intersecting.
b) All angles are right angles: In a square, each of the four angles measures $$90^\circ$$, defining the right angle between adjacent sides.
c) All sides are equal: A distinctive feature of a square is that all four of its sides have the same length, which is key in differentiating it from other types of quadrilaterals.
d) The diagonals bisect each other at right angles: The intersection point of the diagonals in a square divides each diagonal into two equal parts, and they intersect each other at an angle of $$90^\circ$$.
These essential properties outline the basic structure of a square and distinguish it from other polygons.
In a quadrilateral $ABCD$, $AB = 7$ cm, $BC = 5$ cm, $AC = 9$ cm, $AD = 6$ cm, $CD = 2$ cm. Which of the following options is true about the construction of quadrilateral $ABCD$?
A. It is possible to draw the quadrilateral since $AD + DC > AC$
B. It is not possible to draw the quadrilateral since $AD + DC < AC$
C. It is possible to draw the quadrilateral since $AD + DC < AC$
D. It is not possible to draw the quadrilateral since $AD + DC > AC$
The correct option is B: It is not possible to draw the quadrilateral since $AD + DC < AC$.
To constitute a valid triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Considering that a quadrilateral can be thought of as comprising two triangles, we need to check the triangle inequality for each potential triangle formed within the quadrilateral.
Given the side lengths:
$AD = 6 \text{ cm}$
$CD = 2 \text{ cm}$
$AC = 9 \text{ cm}$
We observe that: $$ AD + CD = 6 \text{ cm} + 2 \text{ cm} = 8 \text{ cm} $$ which is less than $AC$ (9 cm).
Since $AD + CD < AC$, the triangle inequality is violated, which implies that no triangle (and hence, no quadrilateral) can be formed with these dimensions. Thus, drawing the quadrilateral $ABCD$ is impossible under these constraints.
Opposite sides are not equal in:
A. Square B. Rectangle C. Rhombus D. Kite
The correct answer is D. Kite.
In geometrical shapes such as a square, rectangle, and rhombus, opposite sides are both equal and parallel. However, in a kite, the opposite sides are generally not equal. This characteristic distinguishes a kite from the other options listed. Thus, the right choice is a Kite.
The area of the trapezium whose vertices lie on the parabola $y^2=4x$ and diagonals pass through $(1,0)$ is $k$ units. If the length of the diagonals is $\frac{25}{4}$ units each, then the value of $4k$ is _____.
To find the value of $4k$, where $k$ is the area of a trapezium formed by vertices on the parabola $y^2 = 4x$, we first analyze the properties of the parabola and the problem's elements:
-
Vertices on the parabola: The parabola given here is $y^2 = 4x$. Each point $(x, y)$ on this parabola follows this equation.
-
Focal properties: For this parabola, considering $4a = 4$, we find $a = 1$. The focus of the parabola is thus at $(1, 0)$. The fact that the diagonals of the trapezium pass through the focus is essential in analyzing the problem.
-
Diagonal characterization: Each diagonal has a length of $\frac{25}{4}$ units.
To solve, we first set up an equation represented by focal chord properties:
-
Since each segment from the focus to a point on the parabola must add up to the whole diagonal, if $AS$ is a segment from the focus to one vertex on the parabola, $CS = \frac{25}{4} - AS$. We also know the formula for segments formed with the focus, $AS = 1 + t^2$, hence $CS = \frac{25}{4} - (1+t^2)$.
-
From the properties of a focal chord, we use reciprocal relationships. For focal chord $AC$: $$ \frac{1}{AS} + \frac{1}{CS} = \frac{1}{a} $$
Using $AS = 1 + t^2$ and solving the equation: $$ \frac{1}{1+t^2} + \frac{4}{25 - 4(1+t^2)} = 1 $$
From which, solving for $t$, using simplifications and factoring, we conclude: $$ 1 + t^2 = 5, \quad \text{or} \quad 1 + t^2 = \frac{5}{4} $$ Thus $t = \pm 2, \pm \frac{1}{2}$. This gives us possible coordinates for vertices A, B, C, and D on the parabola: $$ A\left(\frac{1}{4}, 1\right), \quad B(4, 4), \quad C(4, -4), \quad D \left(\frac{1}{4}, -1\right) $$
The length between $AD$ is $2$ units and $BC$ is $8$ units, with a separation (height of the trapezium) of $\frac{15}{4}$ units.
Calculating the area of the trapezium, $ABCD$: $$ k = \frac{1}{2} \left(AD + BC\right) \times \text{height} = \frac{1}{2} \times (2 + 8) \times \frac{15}{4} = \frac{75}{4} \text{ square units} $$
Therefore, $\boldsymbol{4k = 75}$.
Each bivalent appears to have four chromatids called:
A tetrads
B polypeptide
C quadrivalents
D Diplotene
The correct options are:
- A tetrads
- C quadrivalents
During the Pachytene stage of prophase, chromosomes pair up and shorten, appearing thicker. At this stage, each bivalent consists of four chromatids, which are referred to as tetrads or quadrivalents.
An isosceles trapezium has an area of $36 , \mathrm{cm}^{2}$. The parallel sides are $12 , \mathrm{cm}$ and $6 , \mathrm{cm}$ respectively. The perimeter of the trapezium is
(A) $28 , \mathrm{cm}$
(B) $32 , \mathrm{cm}$
(C) $24 , \mathrm{cm}$
(D) $20 , \mathrm{cm}$
The correct answer is (A) $28 , \mathrm{cm}$.
To find the perimeter of the trapezium, we start by determining its height using the formula for the area of a trapezium: $$ \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} $$ Given: $$ 36 = \frac{1}{2} \times (12 + 6) \times h $$ Solving for $h$: $$ 36 = \frac{1}{2} \times 18 \times h \Longrightarrow h = \frac{36 \times 2}{18} = 4 , \mathrm{cm} $$ Thus, the height $h$ of the trapezium is 4 cm.
Let's now find the lengths of the non-parallel sides (the legs) of the trapezium. For an isosceles trapezium, the legs are equal. Denote the length of each leg as $d$.
Applying the Pythagorean theorem in one of the right triangles formed by the height, a leg, and half the difference of the parallel sides (each 3 cm, since $\frac{12 - 6}{2} = 3$ cm): $$ h^2 + (3 , \mathrm{cm})^2 = d^2 \Longrightarrow 4^2 + 3^2 = d^2 \Longrightarrow 16 + 9 = d^2 \Longrightarrow d^2 = 25 \Longrightarrow d = 5 , \mathrm{cm} $$ This calculation confirms both legs $d = c = 5 , \mathrm{cm}$ are equal.
Finally, the perimeter of the trapezium is the sum of all sides: $$ \text{Perimeter} = 12 , \mathrm{cm} + 6 , \mathrm{cm} + 5 , \mathrm{cm} + 5 , \mathrm{cm} = 28 , \mathrm{cm} $$
Therefore, the perimeter of the trapezium is $28 , \mathrm{cm}$.
If two adjacent vertices of a parallelogram are $(3,2)$ and $(-1,0)$ and the diagonals intersect at $(2,-5)$, then the other two vertices are
A $(1,-10), (5,-12)$
B $(1,-12), (5,-10)$
C $(2,-10), (5,-12)$
D $(1,-10), (2,-12)$
The correct option is B $$ (1,-12), (5,-10) $$
Let vertices $A$ and $B$ of the parallelogram be located at $(3, 2)$ and $(-1, 0)$ respectively. Denote the unknown vertices as $C(x_1, y_1)$ and $D(x_2, y_2)$. Given that the diagonals intersect at $(2, -5)$, we utilize the property that the diagonals of a parallelogram bisect each other.
To find the coordinates of vertices $C$ and $D$, first consider diagonal $AC$. The midpoint of $AC$ should be at the intersection point of diagonals, $(2, -5)$. Using the midpoint formula: $$ \left( \frac{x_1 + 3}{2}, \frac{y_1 + 2}{2} \right) = (2, -5) $$ Solving, we find: $$ \frac{x_1 + 3}{2} = 2 \Rightarrow x_1 = 1 \ \frac{y_1 + 2}{2} = -5 \Rightarrow y_1 = -12 $$ Thus, the coordinates for $C$ are $(1, -12)$.
Next, consider diagonal $BD$ with its midpoint also at $(2, -5)$. Applying the midpoint formula: $$ \left( \frac{x_2 - 1}{2}, \frac{y_2 + 0}{2} \right) = (2, -5) $$ Solving, we get: $$ \frac{x_2 - 1}{2} = 2 \Rightarrow x_2 = 5 \ \frac{y_2 + 0}{2} = -5 \Rightarrow y_2 = -10 $$ Hence, the coordinates for $D$ are $(5, -10)$.
Thus, the coordinates of the other two vertices $C$ and $D$ are $(1, -12)$ and $(5, -10)$ respectively, corresponding to option B.
In parallelogram $ABCD$, $AL \perp CD$ and $CM \perp AD$. If $AL = 20$ cm, $CD = 12$ cm, and $CM = 8$ cm, then the perimeter of the parallelogram is
(A) $42$ cm
(B) $84$ cm
(C) $64$ cm
(D) $68$ cm
The correct answer is (B) $84$ cm.
In parallelogram $ABCD$, since $AL \perp CD$ and $CM \perp AD$, $AL$ and $CM$ are the heights of the parallelogram corresponding to bases $CD$ and $AD$, respectively.
- $AL = 20$ cm is the height from $A$ to $CD$.
- $CD = 12$ cm is the length of base $CD$.
- $CM = 8$ cm is the height from $C$ to $AD$.
To find the length of side $AD$, we use the relationship between the base, height, and area. Since $CD$ and $AD$ are parallel and both $AL$ and $CM$ are perpendicular to their respective bases, both pairs of opposite sides of this parallelogram are equal: $AD = BC$ and $AB = CD$.
The area $A$ of parallelogram $ABCD$ can be calculated as: $$ A = \text{base} \times \text{height} = CD \times AL = 12 , \text{cm} \times 20 , \text{cm} = 240 , \text{cm}^2. $$
Alternatively, using other base-height pair, $$ A = AD \times CM. $$
Since the area $A$ does not change: $$ 240 , \text{cm}^2 = AD \times 8 , \text{cm}, $$
we find $AD$ as: $$ AD = \frac{240 , \text{cm}^2}{8 , \text{cm}} = 30 , \text{cm}. $$
Thus, $BC = 30$ cm as well. Now, knowing all sides:
- $CD = 12$ cm,
- $AD = 30$ cm,
- $AB = 12$ cm,
- $BC = 30$ cm,
the perimeter $P$ of parallelogram $ABCD$ is $$ P = AB + BC + CD + AD = 12 , \text{cm} + 30 , \text{cm} + 12 , \text{cm} + 30 , \text{cm} = 84 , \text{cm}. $$
Therefore, the perimeter of $ABCD$ is 84 cm, confirming option (B) is correct.
A rectangle $ABCD$, where $A=(0,0)$, $B=(4,0)$, $C=(4,2)$, $D=(0,2)$, undergoes the following transformations successively:
(i) $f_{1}(x, y) \rightarrow (y, x)$ (ii) $f_{2}(x, y) \rightarrow (x+3y, y)$ (iii) $f_{3}(x, y) \rightarrow \left( \frac{x-y}{2}, \frac{x+y}{2} \right)$
The final figure will be:
A) a square
B) a rhombus
C) a rectangle
D) a parallelogram
Solution The correct option is D) a parallelogram.
-
Point $A$ remains unchanged at $(0,0)$.
-
For point $B$: $$B = (4,0) \xrightarrow{f_1} (0,4) \xrightarrow{f_2} (12,4) \xrightarrow{f_3} \left(\frac{12-4}{2}, \frac{12+4}{2}\right) = (4,8)$$ So, $B$ finally becomes $(4,8)$.
-
For point $C$: $$C = (4,2) \xrightarrow{f_1} (2,4) \xrightarrow{f_2} (14,4) \xrightarrow{f_3} (5,9)$$
-
For point $D$: $$D = (0,2) \xrightarrow{f_1} (2,0) \xrightarrow{f_2} (2,0) \xrightarrow{f_3} (1,1)$$
Calculating the slopes:
- Slope of $AB = \frac{8-0}{4-0} = 2$
- Slope of $CD = \frac{9-1}{5-1} = 2$
- Slope of $BC = \frac{9-8}{5-4} = 1$
- Slope of $AD = \frac{1-0}{1-0} = 1$
As the slopes indicate,
- $AB$ is parallel to $DC$,
- $BC$ is parallel to $AD$.
Since opposite sides are parallel but the lengths of $AB \neq BC$, and $AB$ is not perpendicular to $BC$, the figure $ABCD$ is a parallelogram.
38 is a multiple of 4.
A) True
B) False
The correct answer is B) False.
To determine if 38 is a multiple of 4, we would expect the remainder to be 0 when 38 is divided by 4.
However, when conducting the division: $$ 38 \div 4 = 9 \text{ remainder } 2 $$ We can express this as: $$ 38 = 4 \times 9 + 2 $$ Since there is a remainder of 2, it indicates that 38 is not a multiple of 4. Thus, the statement is false.
47 The angles $P, Q, R$ and $S$ of a quadrilateral are in the ratio $1: 3: 7: 9$. Then, $P Q R S$ is a a) Parallelogram b) Trapezium with $PQ | RS$ c) Trapezium with $QR | PS$ d) Kite
Solution
We begin by denoting the angles $P, Q, R,$ and $S$ as $x, 3x, 7x,$ and $9x$ respectively. Knowing that the sum of the angles in any quadrilateral is $360^\circ$, we can write the equation: $$ x + 3x + 7x + 9x = 360^\circ $$
Simplifying it, we get: $$ 20x = 360^\circ \ x = \frac{360^\circ}{20} \ x = 18^\circ $$
Using the value of $x$, the angles are assigned as follows:
- $P = 18^\circ$
- $Q = 54^\circ$
- $R = 126^\circ$
- $S = 162^\circ$
Next, we consider the relationships:
- $\angle P + \angle S = 18^\circ + 162^\circ = 180^\circ$
- $\angle Q + \angle R = 54^\circ + 126^\circ = 180^\circ$
The relationships above show opposite angles in pairs summing to $180^\circ$, which indicates that $PQ$ is parallel to $SR$. Therefore, the quadrilateral $PQRS$ is classified as a trapezium with $PQ | SR$.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Solution
Consider a cyclic quadrilateral $ABCD$ with $O$ as the center of its circumscribed circle.
The key concept here is that the perpendicular bisectors of any chord in a circle pass through the center of the circle. Hence, for our cyclic quadrilateral $ABCD$, the perpendicular bisectors of the sides $AB$, $BC$, $CD$, and $DA$ must also pass through $O$, the center of the circle.
These perpendicular bisectors are drawn from $O$ to each side of the quadrilateral. Let us denote them by $OE$, $OF$, $OG$, and $OH$, here $E$, $F$, $G$, and $H$ are the midpoints of sides $AB$, $BC$, $CD$, and $DA$, respectively.
Thus, the perpendicular bisectors $OE$, $OF$, $OG$, and $OH$ are all concurrent at point $O$, proving that the perpendicular bisectors of the sides of a cyclic quadrilateral intersect at a single point, the center of the circumscribed circle.
If the area of parallelogram $ABCD$ is $25 \text{ cm}^{2}$ and on the same base $CD$, a triangle $BCD$ is given such that the area of $BCD$ is $x \text{ cm}^{2}$, then the value of $x$ is:
(A) $12.5 \text{ cm}^{2}$
(B) $25 \text{ cm}^{2}$
(C) $20 \text{ cm}^{2}$
(D) $15 \text{ cm}^{2}$
The correct option is (A) $12.5 \text{ cm}^{2}$.
Explanation: The area of parallelogram $ABCD$ is $25 \text{ cm}^{2}$.
We know that the area of a triangle formed by connecting any vertex to the opposite side of a parallelogram is half the area of the parallelogram. Here, triangle $BCD$ is formed on the same base $CD$ and between the same parallels $CD$ and $AB$ as parallelogram $ABCD$.
Thus, the area of triangle $BCD$ is half of the area of parallelogram $ABCD$. Therefore,
$$ \text{Area of } \triangle BCD = \frac{1}{2} \times 25 \text{ cm}^2 = 12.5 \text{ cm}^2 $$
This makes option (A) the correct answer.
The interior angles of a square are
A. concurrent
B. not concurrent
C. congruent
D. not congruent
The correct answer is C. congruent.
This is because all interior angles of a square are right angles (90 degrees), making them congruent to each other.
The area of a trapezium is $440 \mathrm{~cm}^{2}$. The lengths of the parallel sides are respectively $10 \mathrm{~cm}$ and $14 \mathrm{~cm}$. Find the distance between them.
Solution
The formula to calculate the area of a trapezium is: $$ \text{Area} = \left(\frac{1}{2}\right) \times (\text{sum of parallel sides}) \times h $$ where $h$ is the height (or the distance between the two parallel sides).
Given the area of the trapezium is $440 , \text{cm}^2$ and the lengths of the parallel sides are $10 , \text{cm}$ and $14 , \text{cm}$, we can plug these values into the formula: $$ 440 = \frac{1}{2} \times (10 + 14) \times h $$
Simplifying the sum of the lengths: $$ 440 = \frac{1}{2} \times 24 \times h $$
Further simplification gives: $$ 440 = 12 \times h $$
Solving for $h$, we find: $$ h = \frac{440}{12} = \frac{110}{3} , \text{cm} $$
Thus, the distance between the parallel sides of the trapezium is $\frac{110}{3} , \text{cm}$.
A flooring tile has the shape of a parallelogram whose base is $24 \text{ cm}$ and the corresponding height is $10 \text{ cm}$. How many such tiles are required to cover a floor of area $1080 \text{ m}^2?$.
Solution
Area of the floor: $$1080 , \text{m}^2$$
Base of the tile: $$24 , \text{cm}$$
Height of the tile: $$10 , \text{cm}$$
First, calculate the area of one tile by multiplying its base and height: $$ \text{Area of one tile} = b \times h = 24, \text{cm} \times 10, \text{cm} = 240, \text{cm}^2 $$
Convert this area from square centimeters to square meters: $$ \text{Area of one tile in square meters} = \frac{240}{10,000} = 0.024, \text{m}^2 $$
To cover the entire floor area of $1080 , \text{m}^2$, calculate how many tiles are needed by dividing the total area by the area of one tile: $$ \text{Number of tiles required} = \frac{1080 , \text{m}^2}{0.024 , \text{m}^2} = 45,000 \text{ tiles} $$
Therefore, 45,000 tiles are required to cover the floor area of $1080 , \text{m}^2$.
If a parallelogram and a rectangle are on the same base and between the same parallels, then the perimeter of a parallelogram is __ than the perimeter of a rectangle.
Given that both a parallelogram and a rectangle are based on the same line segment and limited by the same parallel lines, one might be curious about how their perimeters compare.
Here is a concise explanation:
- Parallelogram BDFE shares the same base, DF, with the rectangle ACFE, and both are enclosed by the same parallels.
- Observing segment lengths in the diagram, it's shown that $AE < BE$ and $FC < DF$. This indicates that the sides of the parallelogram (BE and DF) opposite the base are typically longer than those of the rectangle (AE and FC).
This results in the parallelogram having a greater perimeter than the rectangle, given that the cumulative length of its sides surpasses that of the rectangular sides encompassing the same area.
To formalize:
- Perimeter of rectangle = 2(AE + FC)
- Perimeter of parallelogram = 2(BE + DF)
Since $AE < BE$ and $FC < DF$, it follows that the perimeter of the parallelogram is greater than that of the rectangle.
$O$ is a point inside a quadrilateral $ABCD$ which isn't at the point of intersection of diagonals. Prove that
$$ OA + OB + OC + OD > AC + BC $$
The given question appears to contain an error in its specification. The correct form of the inequality that fits the context of geometry involving a point and quadrilateral sides generally should be:
$$ OA + OB + OC + OD > AC + BD $$
where $AC$ and $BD$ are the diagonals of the quadrilateral $ABCD$. It seems there was confusion with the sides and diagonals in the original question's text.
Let's proceed with the corrected inequality, which states that the sum of the distances from a point $O$ inside quadrilateral $ABCD$ to each of its vertices is greater than the sum of its diagonals.
Why the corrected inequality holds: Without loss of generality, consider two separate triangles formed by $O$, one $\triangle OAC$ and the other $\triangle OBD$. By the triangle inequality:
- In $\triangle OAC$, $OA + OC > AC$,
- In $\triangle OBD$, $OB + OD > BD$.
Adding these two inequalities: $$ (OA + OC) + (OB + OD) > AC + BD $$ This simplifies to: $$ OA + OB + OC + OD > AC + BD $$
This is the corrected form and proof based on the application of the triangle inequality theorem.
Construction of a unique quadrilateral is possible if we have the measurement of 3 sides and included angle(s).
A) 3
B) 2
C) 1
D) 4
The correct answer is B) 2.
To construct a unique quadrilateral, we need a total of 5 measurements. Consequently, if we know three sides and two included angles, we can uniquely determine the shape of the quadrilateral. These measurements provide enough information to fix the overall geometry, avoiding any ambiguity in the structure of the quadrilateral. Thus, knowing the measurements of three sides and two included angles is essential.
Which of the following are true? i) All squares are rhombuses. ii) All parallelograms are rectangles. iii) All rhombuses are trapeziums. iv) All squares are rectangles.
A ii, iii
B i, ii
C i, iii, iv
D None of these
Solution
The correct option is D: None of these.
-
i) All squares are rhombuses. A rhombus is defined by having all sides of equal length and the diagonals bisect each other at right angles. A square meets both of these criteria because it has all sides equal and its diagonals bisect each other at $90^\circ$. Therefore, (i) is true.
-
ii) All parallelograms are rectangles. A rectangle is characterized by all angles being $90^\circ$. However, a parallelogram only requires opposite sides to be parallel and equal, without any condition on angles being right angles. Consequently, (ii) is false.
-
iii) All rhombuses are trapeziums. A rhombus has all sides equal and diagonals that bisect each other. A trapezium, in general, is defined by having at least one pair of parallel sides. Since a rhombus does not inherently satisfy this definition singularly by its properties, (iii) is false.
-
iv) All squares are rectangles. A rectangle has all angles at $90^\circ$ and opposite sides equal. A square fulfills both these properties, with the added condition that all four sides are equal. Thus, (iv) is true.
Given the evaluation above, options i and iv are true, but options ii and iii are false, making the correct answer D: None of these.
In quadrilateral $ABCD$, the diagonals intersect at $O$ and $AB | CD$. The areas of triangles $AOB$ and $DOC$ are equal and the altitude of $\triangle AOB$ is $5$ cm. The area of triangle $ABC$ is:
(A) $45$ cm$^{2}$
(B) $75$ cm$^{2}$
(C) $90$ cm$^{2}$
(D) $105$ cm$^{2}$
The correct answer is (A) $45$ cm$^{2}$.
Given that quadrilateral $ABCD$ has diagonals intersecting at $O$ and $AB | CD$, we know the triangles $\triangle AOB$ and $\triangle DOC$ share a similar alignment due to the parallel lines. Let $y$ represent the altitude of $\triangle AOB$, and let $x$ be the altitude of $\triangle DOC$.
Given $\text{Area}(\triangle AOB) = \text{Area}(\triangle DOC)$ and the altitude of $\triangle AOB$ is given to be $y = 5$ cm: $$ \text{Area}(\triangle AOB) = \frac{1}{2} \times \text{base} \times y = \frac{1}{2} \times 6 \times 5 $$ And the corresponding area of $\triangle DOC$ would also be: $$ \text{Area}(\triangle DOC) = \frac{1}{2} \times \text{base} \times x $$ If $\text{Area}(\triangle DOC) = \text{Area}(\triangle AOB)$, then $\frac{1}{2} \times 6 \times 5 = \frac{1}{2} \times 3 \times x$. Solving for $x$, we find: $$ 30 = 3x \Rightarrow x = 10 , \text{cm} $$ Calculating the area of $\triangle ABC$: $$ \text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base of } ABC \times \text{(altitude }y \text{ + altitude }x\text{)} $$ Base of $ABC = 6$ cm as $AB$ and $CD$ are parallel and share similar horizontal alignment: $$ \text{Area}(\triangle ABC) = \frac{1}{2} \times 6 \times (5 + 10) = 45 , \text{cm}^{2} $$ Thus, the area of triangle $ABC$ is $45$ cm$^{2}$, which corresponds to option (A).
If the side of a rhombus is $6 \mathrm{~cm}$ and its one diagonal is $8 \mathrm{~cm}$, find the area of the rhombus in $\mathrm{cm}^{2}$.
A) $9 \sqrt{3}$ B) $3 \sqrt{9}$ C) $16 \sqrt{5}$ D) $8 \sqrt{5}$
To find the area of the rhombus, we use the given information: the side length $s = 6 , \text{cm}$ and one diagonal $d_1 = 8 , \text{cm}$.
Steps:
-
Divide the rhombus into two triangles using the given diagonal.
-
Calculate the semi-perimeter of one of these triangles. The semi-perimeter formula is: $$ \text{semi-perimeter (s)} = \frac{(s + s + d_1)}{2} = \frac{(6 + 6 + 8)}{2} = 10 , \text{cm} $$
-
Use Heron’s formula to calculate the area of one triangle: $$ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} $$ where $a$, $b$, $c$ are the sides of the triangle. Here, $a = b = 6 , \text{cm}$ (sides of the rhombus), $c = 8 , \text{cm}$ (diagonal). $$ \text{Area} = \sqrt{10(10-6)(10-6)(10-8)} = \sqrt{10 \times 4 \times 4 \times 2} = \sqrt{320} = 8\sqrt{5} , \text{cm}^2 $$
-
Double this area to find the area of the rhombus, since the rhombus consists of two such triangles: $$ \text{Area of the rhombus} = 2 \times \text{Area of one triangle} = 2 \times 8 \sqrt{5} = 16 \sqrt{5} , \text{cm}^2 $$
Conclusion: The area of the rhombus is $16 \sqrt{5} , \text{cm}^2$. Thus, the correct answer is $\mathbf{C}$.
In each of the questions below, four statements are given followed by four conclusions numbered I, II, III, and IV. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements, disregarding commonly known facts.
Statements: All books are diaries. Some diaries are pens. Some pens are drawers. All drawers are chairs.
Conclusions: I. Some drawers are diaries. II. Some chairs are pens. III. Some pens are books. IV. Some diaries are books.
Options: A) None follows B) Only II follows C) Only II and III follow D) Only II and IV follow E) All follow
Solution
Statements:
- All books are diaries.
- Some diaries are pens.
- Some pens are drawers.
- All drawers are chairs.
Conclusions:
- Some drawers are diaries. - Not necessarily true as the connection between drawers (which are some pens) and diaries cannot be confirmed directly.
- Some chairs are pens. - True, as all drawers are chairs, and some pens are drawers, thus indirectly some chairs must be pens.
- Some pens are books. - Cannot be confirmed since "some diaries are pens" does not necessarily mean these pens are part of the books subset.
- Some diaries are books. - True, by the definition given in the statement "All books are diaries," it is valid to conclude that some diaries are indeed books as 'all' imply 'some.'
Conclusion based on above analysis: Only conclusions II (Some chairs are pens) and IV (Some diaries are books) logically follow based on the given statements.
Hence, the correct answer is D) Only II and IV follow.
If a circle cuts the rectangular hyperbola $xy = 1$ in the points $(x_{r}, y_{r}), r = 1,2,3,4$, then $x_{1}x_{2}x_{3}x_{4} = y_{1}y_{2}y_{3}y_{4}$ is equal to
A) 1 B) 2 C) 3 D) 4
Solution
The correct option is A) $1$.
Let the equation of the circle be: $$ x^2 + y^2 + 2gx + 2fy + k = 0 $$ Given the equation of the hyperbola: $$ xy = c^2 $$ Now, eliminating $y$ between these two equations, we can substitute $y = \frac{1}{x}$ (since $c^2 = 1$ simplifies our hyperbola equation to $xy = 1$). We then get: $$ x^2 + \frac{1}{x^2} + 2gx + 2f\left(\frac{1}{x}\right) + k = 0 $$ Multiplying through by $x^2$ to clear the fraction, we arrive at a quartic equation: $$ x^4 + 2gx^3 + kx^2 + 2fx + 1 = 0 $$ This equation gives us four roots for $x$, denoted $x_1, x_2, x_3,$ and $x_4$. Hence, the product of these roots, given by the constant term of the equation (as the coefficient of the highest power of $x$ is $1$), is: $$ x_1 x_2 x_3 x_4 = 1 $$
Corresponding to each value of $x_r$, there is a value of $y_r$ such that $x_r y_r = 1$. Thus, we have: $$ y_1 y_2 y_3 y_4 = \frac{1}{x_1 x_2 x_3 x_4} = 1 $$ Therefore, $x_1x_2x_3x_4 = y_1y_2y_3y_4 = 1$.
The points $A(4,-2)$, $B(7,2)$, $C(0,9)$, and $D(-3,5)$ form a parallelogram. Find the length of the altitude on the base $AB$.
To find the length of the altitude, $ h $, on the base $\overline{AB}$ of the parallelogram, start by calculating the length of $\overline{AB}$ itself:
$$ AB = \sqrt{(7-4)^2 + (2+2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ units} $$
The area of the parallelogram can be found using the determinant formula for a polygon with vertices (x, y):
$$ \text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right| $$
Plugging the coordinates of points $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, and $\mathbf{D}$:
$$ \text{Area} = \frac{1}{2} \left| 4 \cdot 2 + 7 \cdot 9 + 0 \cdot 5 - (-3) \cdot (-2) - (-2) \cdot 7 - 2 \cdot 0 - 9 \cdot (-3) \right| $$
$$ \text{Area} = \frac{1}{2} \left| 8 + 63 + 0 + 6 - (-14) - 0 + 27 \right| $$
$$ \text{Area} = \frac{1}{2} \left| 104 \right| = 52 \text{ sq units} $$
Since a parallelogram's area can also be computed as:
$$ \text{Area} = \text{base} \times \text{height} $$
Set this equal to the determinant-based area calculation:
$$ 52 = 5 \times h $$
Finally, solve for $ h $:
$$ h = \frac{52}{5} = 10.4 \text{ units} $$
Therefore, the length of the altitude on the base $ AB $ is $10.4$ units.
In which of the following figures are the sides always congruent line segments?
A Equiangular triangle
B Rhombus
C Square
D All of the above.
The correct option is D All of the above.
-
Equiangular triangle: An equiangular triangle is a type of triangle where all three angles are equal. In a triangle, if all angles are equal, each angle measures $60^\circ$, making the triangle equilateral automatically. This ensures that all sides are congruent.
-
Rhombus: By definition, a rhombus is a quadrilateral where all four sides are of equal length. Thus, all sides are congruent.
-
Square: A square is a specific type of rhombus where not only are all four sides equal in length, but all angles are also right angles. As in a rhombus, all sides are congruent in a square.
Therefore, all the given figures have congruent line segments.
The adjacent angles of a parallelogram add up to:
A) $90^{\circ}$
B) $180^{\circ}$
C) $270^{\circ}$
D) $360^{\circ}$
Solution
The correct answer is B) $180^{\circ}$.
In any parallelogram, such as parallelogram $ABCD$, the properties of the angles are key in answering this question:
- Opposite angles are equal, i.e., $\angle A = \angle C$ and $\angle B = \angle D$.
- Adjacent angles are supplementary, which means: $$ \angle A + \angle B = 180^{\circ} $$ Therefore, in a parallelogram, adjacent angles always add up to $180^{\circ}$.
If the adjacent sides of a parallelogram are represented by $2x^{2} - 5xy + 3y^{2} = 0$ and the equation of one diagonal is $x + y - 2 = 0$, then the equation of the other diagonal is
A) $9x - 11y = 0$
B) $9x + 11y = 0$
C) $11x - 9y = 0`
D) $11x + 9y = 0`
To find the equation of the other diagonal in the given parallelogram, let's first analyze the given equations and use the provided values to establish necessary relationships.
-
Equation of Adjacent Sides: Given the equation $$ 2x^2 - 5xy + 3y^2 = 0, $$ we can factorize it as: $$ 2x^2 - 2xy - 3xy + 3y^2 = 0 \ \Rightarrow (x-y)(2x - 3y) = 0. $$ Hence, the equations of the adjacent sides are $y = x$ and $y = \frac{2x}{3}$ respectively.
-
Given Equation of One Diagonal: The diagonal $AC$ has the equation $$ x + y - 2 = 0. $$
-
Calculating Vertex Coordinates: The origin, $O$, can be assumed where the two adjacent sides meet. Thus:
- For side $OA$ with $y=x$, point $A$ on diagonal $AC$ needs to fulfill $x+y-2=0$. Setting $y=x$, we find $A$ as: $$ x + x - 2 = 0 \ \Rightarrow 2x - 2 = 0 \ \Rightarrow x = 1 \ \Rightarrow A = (1,1). $$
- For side $OC$ with $y=\frac{2x}{3}$ and using point $C$ on diagonal $AC$, setting $y=\frac{2x}{3}$ we get: $$ x + \frac{2x}{3} - 2 = 0 \ \Rightarrow \frac{5x}{3} - 2 = 0 \ \Rightarrow 5x = 6 \ \Rightarrow x = \frac{6}{5}, y = \frac{4}{5} \ \Rightarrow C = \left(\frac{6}{5}, \frac{4}{5}\right). $$
-
Finding the Other Diagonal: With vertices $A$ and $C$ known, the parallelogram's fourth vertex (opposite to origin), $B$, can be derived through symmetry principles. Vertex $B$ will lie on both extended lines from $A$ to $C$ and vice versa. Hence, slope of $OB$ (assuming diagonal $BD$ passes through the origin): $$ \text{Slope} = \frac{\frac{9}{5}}{\frac{11}{5}} = \frac{9}{11}. $$ Therefore, the equation of $OB$ is: $$ y = \frac{9}{11}x. $$ Rewriting in standard form: $$ 11y = 9x \ \Rightarrow \mathbf{9x - 11y = 0}. $$
Conclusion: The equation of the other diagonal is $\mathbf{9x - 11y = 0}$, corresponding to option A.
The sum of the measures of angles of a quadrilateral and that of a triangle is
A) $180^{\circ}$
B) $360^{\circ}$
C) $540^{\circ}$
D) $720^{\circ}$
The correct answer is C) $540^{\circ}$.
To solve this, consider the properties of geometric shapes:
- The sum of the interior angles of a quadrilateral totals $360^{\circ}$.
- The sum of the interior angles of a triangle is $180^{\circ}$.
To find the total sum of the angles for both a quadrilateral and a triangle, we simply add these sums:
$$ 360^{\circ} + 180^{\circ} = 540^{\circ} $$
Thus, the combined sum of the angles is $540^{\circ}$.
If the diagonals of a rhombus are $24 \text{ cm}$ and $10 \text{ cm}$, then its perimeter is $\text{cm}$
A) 48
B) 52
C) 60
D) 80
The correct answer is Option B) $52$ cm.
Given a rhombus where the lengths of the diagonals are:
- $AC = 24$ cm
- $BD = 10$ cm
Each diagonal in a rhombus bisects the other at right angles (i.e., $90^\circ$), dividing the rhombus into four right triangles. Thus,
- Half of $AC$, $AO = OC = \frac{24}{2} = 12$ cm
- Half of $BD$, $BO = OD = \frac{10}{2} = 5$ cm
Consider one of these right triangles, $\triangle AOB$:
- Using the Pythagorean Theorem: $$ AB^2 = AO^2 + BO^2 = 12^2 + 5^2 = 144 + 25 = 169 $$ $$ AB = \sqrt{169} = 13 \text{ cm} $$
Since all sides of a rhombus are equal:
- The perimeter is given by $4 \times AB = 4 \times 13 = 52$ cm.
Thus, the perimeter of the rhombus is 52 cm.
Pavan says the letter D is a polygon. Is he correct?
A) True
B) False
The correct answer is B) False.
A polygon is defined as a 2-dimensional shape that is formed with straight lines only. The letter D comprises one straight line and one curve.
Therefore, the letter D is not a polygon.
Let $ABCD$ be a parallelogram whose diagonals intersect at $P$ and let $O$ be the origin. Then $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}$ equals
(A) $\overrightarrow{OA}$
(B) $2 \overrightarrow{OP}$
(C) $3 \overrightarrow{OP}$
(D) $4 \overrightarrow{OP}$
The correct choice is (D) $4 \overrightarrow{OP}$. In a parallelogram, the diagonals bisect each other, which means that point $P$ is the midpoint of both diagonals $AC$ and $BD$.
Thus, we can write the equations for the midpoint in terms of vector summation: $$ \overrightarrow{OA} + \overrightarrow{OC} = 2\overrightarrow{OP} \quad \text{and} \quad \overrightarrow{OB} + \overrightarrow{OD} = 2\overrightarrow{OP} $$
Adding these equations together provides: $$ (\overrightarrow{OA} + \overrightarrow{OC}) + (\overrightarrow{OB} + \overrightarrow{OD}) = 2\overrightarrow{OP} + 2\overrightarrow{OP} = 4\overrightarrow{OP} $$
Hence, the expression for the sum of vectors from the origin $O$ to each vertex of the parallelogram is: $$ \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} = 4 \overrightarrow{OP} $$
The triangle QTR and the parallelogram PQSR have the same base QR and lie between:
A) PS, QR which are not parallel
B) PS, QR which are parallel
C) PQ, SR which are parallel
D) PR, QS which are not parallel.
The correct option is B) PS, QR which are parallel.
Triangle QTR and the parallelogram PQSR share the same base, $ QR $, and are positioned between the same pair of lines, $ PS $ and $ QR $.
It is specified that $ PQSR $ is a parallelogram. In any parallelogram, opposite sides are parallel. Therefore, $ PS $ and $ QR $ are parallel to each other.
A flooring tile has the shape of a parallelogram whose base is $24 \mathrm{~cm}$ and the corresponding height is $10 \mathrm{~cm}$. The number of such tiles required to cover a floor area of $1080 \mathrm{~m}^{2}$ is:
A) 34000
B) 44000
C) 45000
D) 54000
To determine the number of tiles required to cover a floor area of $1080 \mathrm{~m}^2$, we first need to calculate the area of a single tile. Each tile is shaped as a parallelogram with a base of $24 \mathrm{~cm}$ and a height of $10 \mathrm{~cm}$:
The area $A$ of a parallelogram is given by: $$ A = \text{base} \times \text{height} = 24 , \text{cm} \times 10 , \text{cm} = 240 , \text{cm}^2 $$
Next, we need to convert the floor area from square meters to square centimeters: $$ 1 , \text{m}^2 = 10,000 , \text{cm}^2 $$ Therefore, $$ 1080 , \text{m}^2 = 1080 \times 10,000 , \text{cm}^2 = 10,800,000 , \text{cm}^2 $$
Now, calculate the number of tiles needed by dividing the total area of the floor by the area of one tile: $$ \text{Number of tiles} = \frac{\text{Total area of the floor}}{\text{Area of one tile}} = \frac{10,800,000 , \text{cm}^2}{240 , \text{cm}^2} = 45,000 $$
Thus, 45,000 tiles are needed to cover the floor area. The correct option is C) 45000.
Refer to the following figure. If we were given the coordinates of A, E, and B and the lengths of ED, DC, and BC, how would you go about finding the area of this figure, pick the easiest way.
Break the figure into ΔAEB and trapezium EBCD
Break the figure into ΔAEB, ΔEAB, and ΔBDC
Break the figure into trapeziums EDAX and AXCB where X is a point on DC such that AX is perpendicular to AX.
Can't say
The correct option is A.
Break the figure into $\triangle AEB$ and trapezium $EBCD$.
Triangle $\triangle AEB$:
Since the coordinates of $A$, $E$, and $B$ are provided, the area of $\triangle AEB$ can be easily calculated using the coordinate geometry formula for the area of a triangle:
$$ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$
Trapezium $EBCD$:
Since the lengths of $ED$, $DC$, and $BC$ are provided, the area of the trapezium can be calculated using the trapezium area formula:
$$ \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height} $$
Here, the parallel sides are $ED$ and $BC$, and the height can be derived or given.
Dividing the figure into these simpler shapes (a triangle and a trapezium) simplifies the calculation significantly. Calculating the area of these sections independently and then summing them up is straightforward.
Why not the other options?
Option B: Dividing into $\triangle AEB$, $\triangle EAB$, and $\triangle BDC$ would require Heron's formula and finding the sides for each triangle, making it cumbersome.
Option C: Dividing into trapeziums $EDAX$ and $AXCB would require dropping a perpendicular from $A$ to $DC and then finding the lengths of $DX$ and $XC. This adds complexity and additional calculations, making it the less efficient method.
Thus, option A is the simplest and most efficient approach.
There is a football field. Its four corner flags are kept in position. A corner flag is shown in the image. Is the quadrilateral representing the field defined? Why?
As depicted in the diagram, a corner flag represents the vertex of a quadrilateral. With the four corner flags correctly positioned, each flag marks one of the vertices of the quadrilateral. Therefore, if all four vertices are given, the quadrilateral is completely defined.
What does it mean to have diagonals bisect each other?
That goes across each other
The point where diagonals meet is the midpoint
The angles are congruent
The full diagonal line is congruent to the other diagonal line
Correct Option: B - The point where diagonals meet is the midpoint.
The term "bisect" signifies dividing something into two equal parts. In the context of diagonals of a geometric shape, it means that the diagonals intersect at a point that divides each diagonal into two equal segments.
A parallelogram of area 48 cm² is divided into two congruent triangles. If the height of the triangle is 4 cm, find the length of the base of the triangle.
Given:
The area of a parallelogram is $ 48 , \text{cm}^2 $.
The parallelogram is divided into two congruent triangles.
To Find:
The length of the base of one triangle.
Step-by-Step :
Area of a Parallelogram: $ \text{Area of parallelogram} = 48 , \text{cm}^2 $
Since the parallelogram is divided into two congruent triangles: $$ \text{Area of parallelogram} = 2 \times \text{Area of one triangle} $$
Therefore: $$ \text{Area of one triangle} = \frac{48 , \text{cm}^2}{2} = 24 , \text{cm}^2 $$
Using the formula for the area of a triangle: $$ \text{Area of triangle} = \frac{1}{2} \times \text{base} \times \text{height} $$
Given that the height of the triangle is $ 4 , \text{cm} $: $$ 24 , \text{cm}^2 = \frac{1}{2} \times \text{base} \times 4 , \text{cm} $$
Simplifying the equation to find the base:
$$[ 24 , \text{cm}^2 = 2 \times \text{base} $$
$$ \text{base} = \frac{24 , \text{cm}^2}{2} = 12 , \text{cm} $$
Conclusion:
The length of the base of the triangle is 12 cm.
If the parallel sides of a trapezium are in the ratio 4:5 and its height is 6 cm and its area is 54 cm², then find the lengths of parallel sides.
A. 4 cm, 5 cm
B. 8 cm, 10 cm
C. 12 cm, 15 cm
D. 16 cm, 20 cm
The correct option is $\mathbf{B}$ 8 cm, 10 cm.
Let's denote the lengths of the parallel sides of the trapezium as $4y$ and $5y$. Given the height is 6 cm, and the area is 54 cm², we can use the formula for the area of a trapezium:
$$ \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height} $$
Substituting the given values into the formula:
$$ 54 = \frac{1}{2} \times (4y + 5y) \times 6 $$
Simplify the expression inside the parentheses:
$$ 54 = \frac{1}{2} \times 9y \times 6 $$
Now, multiply inside the parentheses:
$$ 54 = 27y $$
To find $y$, divide both sides by 27:
$$ y = \frac{54}{27} = 2 $$
Hence, the lengths of the parallel sides are:
$$ 4y = 4 \times 2 = 8 \text{ cm} $$ and $$ 5y = 5 \times 2 = 10 \text{ cm} $$
Every parallelogram is a rectangle. True/False
False
In a parallelogram, all opposite sides are parallel and equal, but the angles are not necessarily right angles. On the other hand, in a rectangle, all angles are equal and are right angles. This key difference between their angles means that not every parallelogram can be classified as a rectangle.
The length and breadth of the rectangle is 12 cm and 5 cm respectively. Find the length of the diagonals.
Given that it is a rectangle, the diagonals will be of equal length.
For the given rectangle, the length (l) is 12 cm and the breadth (w) is 5 cm.
Using the Pythagorean theorem to find the length of the diagonal (d), we have:
$$ d^2 = l^2 + w^2 $$
Substituting the given values:
$$ d^2 = 12^2 + 5^2 $$
Calculating the squares:
$$ d^2 = 144 + 25 $$
Adding these values:
$$ d^2 = 169 $$
Taking the square root of both sides to find the length of the diagonal:
$$ d = \sqrt{169} = 13 , \text{cm} $$
Thus, the length of each diagonal is 13 cm.
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.
Given:
$AC = 22$ cm
$BM = 3$ cm
$DN = 3$ cm
$BM \perp AC$
$DN \perp AC$
To find the area of the quadrilateral ABCD, we need to use the formula for the area of triangles and sum the areas of $\triangle ABC$ and $\triangle ADC$.
The formula for the area of a triangle is: [ \text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height} ]
Calculate the area of $\triangle ABC$: [ \text{Area of } \triangle ABC = \frac{1}{2} \times AC \times BM = \frac{1}{2} \times 22 \text{ cm} \times 3 \text{ cm} = \frac{1}{2} \times 66 = 33 \text{ cm}^2 ]
Similarly, calculate the area of $\triangle ADC$: [ \text{Area of } \triangle ADC = \frac{1}{2} \times AC \times DN = \frac{1}{2} \times 22 \text{ cm} \times 3 \text{ cm} = \frac{1}{2} \times 66 = 33 \text{ cm}^2 ]
Summing these two areas gives the total area of quadrilateral ABCD: [ \text{Total Area of Quadrilateral ABCD} = \text{Area of } \triangle ABC + \text{Area of } \triangle ADC ] [ = 33 \text{ cm}^2 + 33 \text{ cm}^2 = 66 \text{ cm}^2 ]
Thus, the area of quadrilateral $ABCD$ is $66 , \text{cm}^2$.
💡 Have more questions?
Ask Chatterbot AI