CBSE Class 10 Mathematics Triangles MCQs Set 09

Objective Type Questions

Question. In a right-angled triangle ABC, right angled at B, \( AB = \frac{x}{2} \), \( BC = x + 2 \) and \( AC = x + 3 \). The value of x is
(a) 5
(b) 10
(c) 12
(d) 14
Answer: (b) 10
Explanation: Here, \( AC^2 = AB^2 + BC^2 \) (by pythagoras theorem)
\( (x + 3)^2 = \left( \frac{x}{2} \right)^2 + (x + 2)^2 \)
This gives, \( x = 10 \)

Question. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. then, the length of the side of the rhombus is:
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Answer: (b) 10 cm
Explanation: 10 cm. A rhombus is a simple quadrilateral whose four sides are of the same length and diagonals are perpendicular bisectors of each other. It is given that AC = 16 cm and BD = 12 cm. \( \angle AOB = 90^\circ \). Since AC and BD bisect each other
\( \implies \) \( AO = \frac{1}{2} AC \) and \( BO = \frac{1}{2} BD \)
\( \implies \) \( AO = 8 \text{ cm} \) and \( BO = 6 \text{ cm} \)
In right angled \(\triangle AOB\),
\( AB^2 = AO^2 + OB^2 \) [Using the Pythagoras theorem]
\( AB^2 = 8^2 + 6^2 = 64 + 36 = 100 \)
\( \implies \) \( AB = \sqrt{100} = 10 \text{ cm} \)

Question. If D, E and F are the mid-points of sides BC, CA and AB respectively of \(\triangle ABC\), then the ratio of the areas of \(\triangle DEF\) to the area of \(\triangle ABC\) is:
(a) 1 : 4
(b) 1 : 2
(c) 1 : 3
(d) 2 : 3
Answer: (a) 1 : 4
Explanation: Here, four small triangles are congruent to each other. So, \( ar(\triangle DEF) = \frac{1}{4} ar(\triangle ABC) \)

Question. If \(\triangle ABC \sim \triangle EDF\) and \(\triangle ABC\) is not similar to \(\triangle DEF\), then which of the following is not true?
(a) \( BC \cdot EF = AC \cdot FD \)
(b) \( AB \cdot EF = AC \cdot DE \)
(c) \( BC \cdot DE = AB \cdot EF \)
(d) \( BC \cdot DE = AB \cdot FD \)
Answer: (c) \( BC \cdot DE = AB \cdot EF \)
Explanation: We know that if sides of one triangle are proportional to the side of the other triangle, and the corresponding angles are also equal, then the triangles are similar by SSS similarity. It is given that \(\triangle ABC \sim \triangle EDF\)
\( \implies \) \( \frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} \)
Also, \(\triangle ABC\) is not similar to \(\triangle DEF\)
\( \frac{AB}{DE} \neq \frac{BC}{EF} \)
i.e., \( AB \cdot EF \neq BC \cdot DE \). Hence (c) is not true.

Question. If in two triangles ABC and PQR, \( \frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ} \), then:
(a) \( \triangle PQR \sim \triangle CAB \)
(b) \( \triangle PQR \sim \triangle ABC \)
(c) \( \triangle CBA \sim \triangle PQR \)
(d) \( \triangle BCA \sim \triangle PQR \)
Answer: (a) \( \triangle PQR \sim \triangle CAB \)
Explanation: It is given that in \(\triangle ABC\) and \(\triangle PQR\), \( \frac{AB}{QR} = \frac{BC}{PR} = \frac{CA}{PQ} \). This shows that the sides of one triangle are proportional to the side of the other triangle, thus their corresponding angles are also equal. i.e., \( \angle A = \angle Q, \angle B = \angle R \) and \( \angle C = \angle P \). Thus, \( \triangle PQR \sim \triangle CAB \).

Question. In \(\triangle ABC\) and \(\triangle DEF\), \( \angle B = \angle E \), \( \angle F = \angle C \) and AB = 3DE. Then, the two triangles are:
(a) congruent but not similar
(b) similar but not congruent
(c) neither congruent nor similar
(d) congruent as well as similar
Answer: (b) similar but not congruent
Explanation: In \(\triangle ABC\) and \(\triangle DEF\)
\( \angle B = \angle E \) [Given]
\( \angle F = \angle C \) [Given]
\( \implies \) \( \triangle ABC \sim \triangle DEF \) [By AA similarity criterion]
AB and DE sides are corresponding sides. But \( AB = 3DE \) [Given].
We know that two triangles are congruent if they have the same shape and size and satisfy the rule of congruency. But \(\triangle ABC\) and \(\triangle DEF\) do not satisfy any rule of congruency, which are SAS, ASA, AAA and SSS, so both are not congruent.
\( \implies \) \(\triangle ABC\) cannot be congruent to \(\triangle DEF\). Hence, \(\triangle\)'s are similar but not congruent.

Question. If in two triangles \(\triangle DEF\) and \(\triangle PQR\), \( \angle D = \angle Q \) and \( \angle R = \angle E \), then which of the following is not true?
(a) \( \frac{EF}{PR} = \frac{DF}{PQ} \)
(b) \( \frac{DE}{PQ} = \frac{EF}{RP} \)
(c) \( \frac{DE}{QR} = \frac{DF}{PQ} \)
(d) \( \frac{EF}{RP} = \frac{DE}{QR} \)
Answer: (b) \( \frac{DE}{PQ} = \frac{EF}{RP} \)
Explanation: It is given that in \(\triangle DEF\) and \(\triangle PQR\), \( \angle D = \angle Q \) and \( \angle R = \angle E \). We know that if two corresponding angles of two triangles are congruent, then both the triangles are similar because if two angle pairs are equal, then the third angle must also be equal.
\( \implies \) \( \triangle DEF \sim \triangle QRP \) [By AA similarity criterion]
\( \implies \) \( \angle F = \angle P \) [Corresponding angles of similar triangles]
\( \frac{DF}{QP} = \frac{ED}{RQ} = \frac{FE}{PR} \). Hence, except option (b), all are true.

Question. It is given that \(\triangle ABC \sim \triangle PQR\), with \( \frac{BC}{QR} = \frac{1}{3} \), then \( \frac{ar(PRQ)}{ar(BCA)} \) is equal to
(a) 9
(b) 3
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{9} \)
Answer: (a) 9
Explanation: It is given that \(\triangle ABC \sim \triangle PQR\). We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{ar(PRQ)}{ar(BCA)} = \left( \frac{QR}{BC} \right)^2 \)
\( \frac{ar(PRQ)}{ar(BCA)} = \left( \frac{3}{1} \right)^2 = \frac{9}{1} \). Thus, the area of \(\triangle PRQ = 9\) times the area of \(\triangle BCA\).

Question. If in triangles ABC and DEF, \( \frac{AB}{DE} = \frac{BC}{FD} \), then they will be similar when:
(a) \( \angle B = \angle E \)
(b) \( \angle A = \angle D \)
(c) \( \angle B = \angle D \)
(d) \( \angle A = \angle F \)
Answer: (c) \( \angle B = \angle D \)
Explanation: In \(\triangle ABC\) and \(\triangle DEF\), \( \frac{AB}{DE} = \frac{BC}{FD} \). Angle formed by AB and BC is \( \angle B \). Angle formed by DE and FD is \( \angle D \).
\( \implies \) \( \angle B = \angle D \). \(\therefore \triangle ABC \sim \triangle DEF\) [By SAS similarity criterion]. Hence, (c) is the correct answer.

Question. If \(\triangle ABC \sim \triangle QRP\), \( \frac{ar(ABC)}{ar(PQR)} = \frac{9}{4} \), AB = 18 cm and BC = 15 cm, then PR is equal to:
(a) 10 cm
(b) 12 cm
(c) \( \frac{20}{3} \) cm
(d) 8 cm
Answer: (a) 10 cm
Explanation: It is given that \(\triangle ABC \sim \triangle QRP\). By similar triangles area property, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.
\( \implies \frac{ar(ABC)}{ar(PQR)} = \frac{BC^2}{PR^2} \)
\( \implies \frac{9}{4} = \frac{BC^2}{PR^2} \). It is given that AB = 18 cm and BC = 15 cm.
\( \implies \frac{15^2}{PR^2} = \frac{9}{4} \)
\( \implies PR^2 = \frac{225 \times 4}{9} = 100 \)
\( \implies PR = 10 \text{ cm} \)

Question. If S is a point on side PQ of a \(\triangle PQR\) such that PS = QS = RS, then:
(a) \( PR \cdot QR = RS^2 \)
(b) \( QS^2 + RS^2 = QR^2 \)
(c) \( PR^2 + QR^2 = PQ^2 \)
(d) \( PS^2 + RS^2 = PR^2 \)
Answer: (c) \( PR^2 + QR^2 = PQ^2 \)
Explanation: In \(\triangle PQR\), PS = QS = RS [Given]. Let \( \angle PRS = \angle 1 \) and \( \angle SRQ = \angle 2 \). In \(\triangle PSR\), PS = RS [Given].
\( \implies \) \( \angle 1 = \angle P \) [Angles opposite to equal sides in a triangle are equal]. Similarly, in \(\triangle SRQ\), RS = SQ [Given].
\( \implies \) \( \angle Q = \angle 2 \). Now, in \(\triangle PQR\), \( \angle P + \angle Q + \angle PRQ = 180^\circ \) [Angle sum property of a triangle].
\( \implies \) \( \angle 1 + \angle 2 + (\angle 1 + \angle 2) = 180^\circ \)
\( \implies \) \( 2(\angle 1 + \angle 2) = 180^\circ \)
\( \implies \) \( \angle 1 + \angle 2 = 90^\circ \)
\( \implies \) \( \angle PRQ = 90^\circ \). By Pythagoras theorem, we have \( PQ^2 = PR^2 + RQ^2 \).

Fill in the Blanks

Question. Let \(\triangle ABC \sim \triangle DEF\) and their areas be 81 \( \text{cm}^2 \) and 144 \( \text{cm}^2 \). If EF = 24 cm, then length of side BC is ...................... cm.
Answer: 18 cm
Explanation: Since \(\triangle ABC \sim \triangle DEF\),
\( \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{BC^2}{EF^2} \)
\( \implies \frac{81}{144} = \frac{BC^2}{24^2} \)
\( \implies BC^2 = \frac{81}{144} \times 24^2 = 324 \)
\( \implies BC = 18 \text{ cm} \)

Question. In \(\triangle ABC\), AB = \( 6\sqrt{3} \) cm, AC = 12 cm and BC = 6 cm, then \(\angle B =\) ...............................
Answer: \(\angle B = 90^\circ\)
Explanation: Since, \( 12^2 = (6\sqrt{3})^2 + 6^2 \), \( 144 = 108 + 36 \).
\( \therefore \) By the converse of the Pythagoras theorem, \( \angle B = 90^\circ \).

Question. Two triangles are similar if their corresponding sides are ...............................
Answer: Proportional

Question. A ladder 10 m long reaches a window 8 m above the ground. The distance of the foot of the ladder from the base of the wall is ................................... m.
Answer: 6 m

Question. If \(\triangle ABC\) is an equilateral triangle of side 2a, then length of one of its altitude is ..............................
Answer: \(\sqrt{3}a\)
Explanation: By the Pythagoras Theorem in \(\triangle ABD\).
Altitude, \( AD = \sqrt{AB^2 - BD^2} \)
\( AD = \sqrt{(2a)^2 - a^2} \)
\( \implies AD = \sqrt{4a^2 - a^2} = \sqrt{3a^2} \)
\( \implies AD = \sqrt{3} \cdot a \)

Question. The perimeters of two similar triangles \(\triangle ABC\) and \(\triangle PQR\) are 35cm and 45cm respectively, then the ratio of the areas of the two triangles is .................... .
Answer: 49 : 81

Question. The length of an altitude in an equilateral triangle of side 'a' cm is ......................... .
Answer: \( \frac{\sqrt{3}}{2}a \) cm
Explanation: Here, \( AB^2 = BD^2 + AD^2 \)
i.e. \( a^2 = \left( \frac{a}{2} \right)^2 + AD^2 \)
\( \implies AD^2 = \frac{3a^2}{4} \)
\( \implies AD = \frac{\sqrt{3}}{2}a \)

Question. If areas of two similar triangles are equal, then these triangles are .................... .
Answer: Congruent

Question. Diagonals of a parallelogram separate it into two triangles of ...................
Answer: equal areas

Question. If S is a point on side PQ of a \(\triangle PQR\) such that PS = QS = RS, then \( PR^2 + QR^2 = \) .................
Answer: \( PQ^2 \)