CBSE Class 10 Mathematics Triangles VBQs Set 02

Multiple Choice 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)
Answer: 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)
Answer: 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 \( AC = 16 \text{ cm} \) and \( BD = 12 \text{ cm} \)
\( \angle AOB = 90^\circ \)
Since AC and BD bisect each other
\( \Rightarrow AO = \frac{1}{2} AC \) and \( BO = \frac{1}{2} BD \)
\( \Rightarrow AO = 8 \text{ cm} \) and \( BO = 6 \text{ cm} \)
In right angled \( \triangle AOB \),
\( AB^2 = AO^2 + OB^2 \) [Using the Pythagoras theorem]
\( AB^2 = 8^2 + 6^2 = 64 + 36 = 100 \)
\( \Rightarrow AB = \sqrt{100} = 10 \text{ cm} \)

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)
Answer: Here, four small triangles are congruent to each other.
So, \( \text{ar}(\triangle DEF) = \frac{1}{4} \text{ar}(\triangle ABC) \)

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)
Answer: 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 \( \triangle ABC \sim \triangle EDF \)
\( \Rightarrow \frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} \)
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)
Answer: 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)
Answer: Explanation:
In \( \triangle ABC \) and \( \triangle DEF \)
\( \angle B = \angle E \) [Given]
\( \angle F = \angle C \) [Given]
\( \Rightarrow \triangle ABC \sim \triangle DEF \) [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.
\( \Rightarrow \triangle ABC \) cannot be congruent to \( \triangle DEF \).
Hence, \( \triangle \)'s are similar but not congruent.

Question. If in two triangles DEF and 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)
Answer: 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.
\( \Rightarrow \triangle DEF \sim \triangle QRP \) [By AA similarity criterion]
\( \Rightarrow \angle F = \angle P \) [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 \( \frac{\text{ar}(PRQ)}{\text{ar}(BCA)} \) is equal to:
(a) 9
(b) 3
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{9} \)
Answer: (a)
Answer: Explanation:
It is given that \( \triangle ABC \sim \triangle PQR \) ...(i)
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
\( \frac{\text{ar}(PRQ)}{\text{ar}(BCA)} = \left(\frac{QR}{BC}\right)^2 \)
\( \frac{\text{ar}(PRQ)}{\text{ar}(BCA)} = \left(\frac{3}{1}\right)^2 = \frac{9}{1} \) [Using eq. (i)]
Thus, the area of \( \triangle PRQ = 9 \) times the area of \( \triangle ABC \).

Trick Applied
Ratio of the area of two similar triangles is equal to the ratio of the squares of their corresponding sides.

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)
Answer: 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 \( \angle D \).
\( \Rightarrow \angle B = \angle D \)
\( \therefore \triangle ABC \sim \triangle DEF \) [By SAS similarity criterion]
Hence, (c) is the correct answer.

Question. If \( \triangle ABC \sim \triangle QRP \), \( \frac{\text{ar}(ABC)}{\text{ar}(PQR)} = \frac{9}{4} \), AB = 18 cm and BC = 15 cm, then PR is equal to:
(a) 10 cm
(b) 12 cm
(c) \( \frac{20}{3} \text{ cm} \)
(d) 8 cm
Answer: (a)
Answer: 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.
\( \Rightarrow \frac{\text{ar}(ABC)}{\text{ar}(PQR)} = \frac{BC^2}{PR^2} \)
\( \frac{9}{4} = \frac{BC^2}{PR^2} \) [Given]
It is given that AB = 18 cm and BC = 15 cm
\( \Rightarrow \frac{15^2}{PR^2} = \frac{9}{4} \)
\( \Rightarrow PR^2 = \frac{225 \times 4}{9} = 100 \)
\( \Rightarrow PR = 10 \text{ cm} \)

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)
Answer: Explanation:
In \( \triangle PQR \), PS = QS = RS [Given]
Let \( \angle PRS = \angle 1 \) and \( \angle SRQ = \angle 2 \)
In \( \triangle PSR \), PS = RS [Given]
\( \Rightarrow \angle 1 = \angle P \) [Angles opposite to equal sides in a triangle are equal]
Similarly, in \( \triangle SRQ \)
RS = SQ [Given]
\( \Rightarrow \angle Q = \angle 2 \)
Now, in \( \triangle PQR \),
\( \angle P + \angle Q + \angle PRQ = 180^\circ \) [Angle sum property of a triangle]
\( \angle 1 + \angle 2 + (\angle 1 + \angle 2) = 180^\circ \)
\( 2(\angle 1 + \angle 2) = 180^\circ \)
\( \angle 1 + \angle 2 = 90^\circ \)
\( \angle PRQ = 90^\circ \)
By Pythagoras Theorem, \( PR^2 + QR^2 = PQ^2 \)

Fill in the Blanks

Fill in the blanks/tables with suitable information:

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

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

Question. Two triangles are similar if their corresponding sides are ............................... 
Answer: Proportional (by definition)

Question. A ladder \(10\text{ m}\) long reaches a window \(8\text{ m}\) above the ground. The distance of the foot of the ladder from the base of the wall is ................................. m. 
Answer: 6 m
From the figure:
By Pythagoras Theorem,
\(AB^2 = AC^2 - BC^2\)
\(= 10^2 - 8^2\)
\(= 100 - 64\)
\(= 36\)
\(\Rightarrow AB = 6\text{ m}\).
Thus, the distance of the foot of the ladder from the base of the wall is \(6\text{ m}\).

Question. If \(\Delta ABC\) 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 \(\Delta ABD\).
Altitude, \(AD = \sqrt{AB^2 - BD^2}\)
\(= \sqrt{(2a)^2 - a^2}\)
\(= \sqrt{4a^2 - a^2}\)
\(= \sqrt{3a^2}\)
\(= \sqrt{3}a\)

Question. The perimeters of two similar triangles \(\Delta ABC\) and \(\Delta PQR\) are \(35\text{cm}\) and \(45\text{cm}\) 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 \(a\text{ cm}\) is ......................... .
Answer: \(\frac{\sqrt{3}}{2}a\text{ cm}\)
Here, \(AB^2 = BD^2 + AD^2\)
i.e. \(a^2 = \left(\frac{a}{2}\right)^2 + AD^2\)
\(\Rightarrow AD^2 = \frac{3a^2}{4} \Rightarrow AD = \frac{\sqrt{3}}{2}a\)

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 \(\Delta PQR\) such that \(PS = QS = RS\), then \(PR^2 + QR^2 = \text{...............}\)
Answer: \(PQ^2\)

Very Short Questions

Question. It is given that \(\Delta DEF \sim \Delta RPQ\). Is it true to say that \(\angle D = \angle R\) and \(\angle F = \angle P\)? Why? 
Answer: No.
Explanation:
We know that when \(\Delta DEF \sim \Delta RPQ\), each angle of triangle \(DEF\) is equal to the corresponding angle of \(\Delta RPQ\). So,
\(\angle D = \angle R; \angle E = \angle P; \angle F = \angle Q\)
\(\Rightarrow \angle D = \angle R\) is true but \(\angle F \neq \angle P\).
Hence, it is not true that \(\angle D = \angle R\) and \(\angle F = \angle P\).

Question. Is the following statement true? Why? "Two quadrilaterals are similar, if their corresponding angles are equal". 
Answer: No. Two quadrilaterals will be similar, if their corresponding angles are equal and ratio of corresponding sides must also be proportional. e.g. : rectangle, square

Question. The ratio of the corresponding altitudes of two similar triangles is \(\frac{3}{5}\). Is it correct to say that the ratio of their areas is \(\frac{6}{5}\)? Why? 
Answer: No.
It is given that the ratio of altitudes of similar triangles is \(\frac{3}{5}\).
Using the property of area of two similar triangles, we have
\(\frac{Area 1}{Area 2} = \left(\frac{Altitude 1}{Altitude 2}\right)^2\)
\(\frac{Area 1}{Area 2} = \left(\frac{3}{5}\right)^2\)
\(\frac{Area 1}{Area 2} = \frac{9}{25}\)
\(\frac{9}{25} \neq \frac{6}{5}\)
Hence, the given statement is not correct because it does not satisfy the criteria.

Question. The area of two similar triangles are \(25\text{ sq. m}\) and \(121\text{ sq. cm}\). Find the ratio of their corresponding sides. 
Answer: Given, \(Area(\Delta_1) = 25\text{ sq. m}\), \(Area(\Delta_2) = 121\text{ sq. cm}\) and \(\Delta_1 \sim \Delta_2\)
Now, \(\frac{ar(\Delta_1)}{ar(\Delta_2)} = \frac{(side_1)^2}{(side_2)^2}\) [\(\because\) ratio of area of similar triangle is equal to ratio of the square of their corresponding sides]
\(\therefore \frac{25}{121} = \left(\frac{side_1}{side_2}\right)^2\)
\(\Rightarrow \frac{side_1}{side_2} = \frac{5}{11}\)
Hence the required ratio is \(5 : 11\).