CBSE Class 10 Mathematics Triangles MCQs Set E

Question. The areas of two similar triangles are \(81 \text{ cm}^2\) and \(49 \text{ cm}^2\) respectively, then the ratio of their corresponding medians is
(a) \(7 : 9\)
(b) \(9 : 81\)
(c) \(9 : 7\)
(d) \(81 : 7\)
Answer: C
Given, area of two similar triangles,
\(A_1 = 81 \text{ cm}^2\)
\(A_2 = 49 \text{ cm}^2\)
Ratio of corresponding medians \(= \sqrt{\frac{A_1}{A_2}} = \sqrt{\frac{81}{49}} = \frac{9}{7}\)

Question. Sides of two similar triangles are in the ratio \(4:9\). Areas of these triangles are in the ratio.
(a) \(2:3\)
(b) \(4:9\)
(c) \(81:16\)
(d) \(16:81\)
Answer: D
We have two similar triangles such that the ratio of their corresponding sides is \(4:9\).
The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\(\frac{\text{ar}(\Delta_1)}{\text{ar}(\Delta_2)} = \left(\frac{4}{9}\right)^2 = \frac{16}{81}\)
\(\text{ar}(\Delta_1) : \text{ar}(\Delta_2) = 16 : 81\)

Question. In a right angled \(\Delta ABC\) right angled at \(B\), if \(P\) and \(Q\) are points on the sides \(AB\) and \(BC\) respectively, then
(a) \(AQ^2 + CP^2 = 2(AC^2 + PQ^2)\)
(b) \(2(AQ^2 + CP^2) = AC^2 + PQ^2\)
(c) \(AQ^2 + CP^2 = AC^2 + PQ^2\)
(d) \(AQ + CP = \frac{1}{2}(AC + PQ)\)
Answer: C

Question. It is given that, \(\Delta ABC \sim \Delta EDF\) such that \(AB = 5 \text{ cm}, AC = 7 \text{ cm}, DF = 15 \text{ cm}\) and \(DE = 12 \text{ cm}\), then the sum of the remaining sides of the triangles is
(a) \(23.05 \text{ cm}\)
(b) \(16.8 \text{ cm}\)
(c) \(6.25 \text{ cm}\)
(d) \(24 \text{ cm}\)
Answer: A

Question. The area of a right angled triangle is \(40 \text{ sq cm}\) and its perimeter is \(40 \text{ cm}\). The length of its hypotenuse is
(a) \(16 \text{ cm}\)
(b) \(18 \text{ cm}\)
(c) \(17 \text{ cm}\)
(d) data insufficient
Answer: B
Suppose hypotenuse of the triangle is \(c\) and other sides are \(a\) and \(b\), obviously.
\(c = \sqrt{a^2 + b^2}\)
We have, \(a + b + c = 40\) and \(\frac{1}{2}ab = 40 \implies ab = 80\)
\(c = 40 - (a + b)\) and \(ab = 80\)
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Question. The area of the semi-circle drawn on the hypotenuse of a right angled triangle is equal
(a) Sum of the areas of the semi-circles drawn on the other two sides of the triangle.
(b) difference of the areas of semi-circles drawn on the other two sides of the triangle.
(c) product of the areas of semi-circles drown on the other two sides of the triangle
(d) None of these
Answer: C

Question. A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6m towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is
(a) 0.6 m
(b) 0.2 m
(c) 0.4 m
(d) 0.8 m
Answer: D

Question. In a \(\Delta PQR\), L and M are two points on base \(QR\), such that \(\angle LPQ = \angle QRP\) and \(\angle RPM = \angle RQP\). Then, which of the following is/are true:
1. \(\Delta PQL \sim \Delta RPM\)
2. \(QL \times RM = PL \times PM\)
3. \(PQ^2 = QR \times QL\)
(a) Both (1) and (2)
(b) Both (2) and (3)
(c) Both (1) and (3)
(d) All the three
Answer: D

Question. \(\Delta ABC\) is right angled at \(A\) with \(AB = 6 \text{ cm}\), \(BC = 10 \text{ cm}\). A circle with centre O has been inscribed inside the triangle. The radius of the incircle is
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 1 cm
Answer: C
2 cm
Given, a right angled \(\Delta ABC\), in which a circle of centre O is inscribed., the sides of a triangle are \(AB = 6 \text{ cm}\), \(BC = 10 \text{ cm}\) and \(AC\). Join \(AO, OB\) and \(OC\).
Now, draw perpendicular from \(O\) to \(AB, BC\) and \(CA\) meeting them at \(D, E\), and \(F\), respectively.
\(BC^2 = AB^2 + AC^2\) [by Pythagoras theorem]
\((10)^2 = (6)^2 + (AC)^2\) [\(BC = 10 \text{ cm}, AB = 6 \text{ cm}\)]
\(100 = 36 + AC^2\)
\(AC^2 = 100 - 36 = 64\)
\(AC = \sqrt{64} = 8 \text{ cm}\)
Now, \(ar(\Delta ABC) = \frac{1}{2} \times AB \times AC\)
\(= \frac{1}{2} \times 6 \times 8 = 24 \text{ sq cm}\) ...(1)
[area of a triangle \(= \frac{1}{2} \times \text{base} \times \text{height}\)]
Also, \(ar(\Delta ABC) = ar(\Delta AOB) + ar(\Delta BOC) + ar(\Delta AOC)\)
\(= \left( \frac{1}{2} \times r \times AB \right) + \left( \frac{1}{2} \times r \times BC \right) + \left( \frac{1}{2} \times r \times AC \right)\)
[\(OD = OE = OF = r\)]
\(= \frac{1}{2} \times r \times (AB + BC + AC)\)
\(= \frac{1}{2} \times r \times (6 + 10 + 8) = 12r\) ...(2)
From Eq. (1) and (2),
\(12r = 24 \Rightarrow r = 2\)
Hence, the radius of the incidence of \(\Delta ABC\) is 2 cm.

FILL IN THE BLANK

Question. A line drawn through the mid-point of one side of a triangle parallel to another side bisects the .......... side.
Answer: third

Question. .......... theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Answer: Pythagoras

Question. Line joining the mid-points of any two sides of a triangle is .......... to the third side.
Answer: parallel

Question. All squares are ..........
Answer: similar

Question. Two triangles are said to be .......... if corresponding angles of two triangles are equal.
Answer: equiangular

Question. All similar figures need not be ..........
Answer: congruent

Question. The ratio of the areas of two similar triangles is equal to the square of the ratio of their ..........
Answer: corresponding sides

Question. Two polygons of the same number of sides are similar, if their corresponding angles are .......... and their corresponding sides are in the same ..........
Answer: equal, ratio

Question. All circles are ..........
Answer: similar

Question. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the .......... side.
Answer: third

Question. If a line divides any two sides of a triangle in the same ratio, then the line is .......... to the third side.
Answer: parallel

TRUE/FALSE

Question. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar.
Answer: True

Question. Two photographs of the same size of the same person at the age of 20 years and the other at the age of 45 years are not similar.
Answer: True

Question. A square and a rectangle are similar figure as each angle of the two quadrilaterals is \(90^\circ\).
Answer: False

Question. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.
Answer: True

Question. If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar.
Answer: True

Question. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Answer: True

Question. All congruent figures need not be similar.
Answer: False

Question. All the congruent figures are similar but the converse is not true.
Answer: True

Question. A circle of radius 3 cm and a square of side 3 cm are similar figures.
Answer: False

Question. In \(\Delta PQR\), if \(X\) and \(Y\) are points on sides \(PQ\) and \(PR\) such that \(\frac{PX}{XQ} = \frac{4}{18}\) and \(\frac{PY}{PR} = \frac{6}{27}\), than \(RQ\) is not parallel to \(XY\).
Answer: True

ASSERTION AND REASON

DIRECTION: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion : Two similar triangle are always congruent.
Reason : If the areas of two similar triangles are equal then the triangles are congruent.
(a) A
(b) B
(c) C
(d) D
Answer: D
Assertion (A) is false but reason (R) is true. Two similar triangles are not congruent generally. So, A is incorrect but R is correct.

Question. Assertion : \(ABC\) is an isosceles, right triangle, right angled at \(C\). Then \(AB^2 = 3AC^2\).
Reason : In an isosceles triangle \(ABC\) if \(AC = BC\) and \(AB^2 = 2AC^2\), then \(\angle C = 90^\circ\).
(a) A
(b) B
(c) C
(d) D
Answer: D

Question. Assertion : \(ABC\) and \(DEF\) are two similar triangles such that \(BC = 4\) cm, \(EF = 5\) cm and area of \(\Delta ABC = 64 \text{ cm}^2\), then area of \(\Delta DEF = 100 \text{ cm}^2\).
Reason : The areas of two similar triangles are in the ratio of the squares of the corresponding altitudes.
(a) A
(b) B
(c) C
(d) D
Answer: B
It both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion. Reason is true [standard result]. For Assertion, since \(\Delta ABC \sim \Delta DEF\), \(\frac{area(\Delta ABC)}{area(\Delta DEF)} = \frac{BC^2}{EF^2} = \frac{4^2}{5^2} = \frac{16}{25}\). (ratio of areas of two similar \(\Delta\)s is equal to the ratio of the squares of corresponding sides). \(\frac{64}{area(\Delta DEF)} = \frac{16}{25} \Rightarrow area(\Delta DEF) = \frac{64 \times 25}{16} = 4 \times 25 = 100 \text{ cm}^2\). Assertion is true. But Reason is not correct explanation for Assertion.

Question. Assertion : In the \(\Delta ABC\), \(AB = 24\) cm, \(BC = 10\) cm and \(AC = 26\) cm, then \(\Delta ABC\) is a right angle triangle.
Reason : If in two triangles, their corresponding angles are equal, then the triangles are similar.
(a) A
(b) B
(c) C
(d) D
Answer: B
Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). We have, \(AB^2 + BC^2 = (24)^2 + (10)^2 = 576 + 100 = 676 = AC^2\). \(ABC\) is a right angled triangle. Also, two triangle are similar if their corresponding angles are equal. So, both A and R are correct but R does not explain A.

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