Practice MCQ Questions and Answer on Triangles

ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true?

(A) PB = PC
(B) PB > PC
(C) PB PC
(D) PB = $\frac{1}{2}$ PC

If the length of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is :

(A) 7.5 cm
(B) 6 cm
(C) 10 cm
(D) 8 cm

Solution: According to question, Area of Δ(OBA + OAC + OBC) = area of ΔABC $\frac{1}{2}$ × 4x × 3 + $\frac{1}{2}$ × 5x × 3 + $\frac{1}{2}$ × 6x × 3 = $\frac{1}{2}$ × (6x × AD) $\frac{1}{2}$ × 3(4x + 5x + 6x) = $\frac{1}{2}$ × (6x × AD) 45x = 6x × AD AD = $\frac{15}{2}$ AD = 7.5 cm

BL and CM are medians of ÃÂÃÂÃÂÃÂÃÂÃÂABC right-angled at A and BC = 5 cm. If BL = $\frac{3 \sqrt{5}}{2}$ cm, then the length of CM is

(A) $2 \sqrt{5}$ cm
(B) $5 \sqrt{2}$ cm
(C) $10 \sqrt{2}$ cm
(D) $4 \sqrt{5}$ cm

Solution: According to question, According to figure, when two medians intersect each other in a right angled triangle then we use this equation. ⇒ 4 (BL2 + CM2) = 5BC2 ⇒ 4 × ${(\frac{3 \sqrt{5}}{2})}^{2}$ + 4CM2 = 5BC2 ⇒ 45 + 4CM2 = 125 ⇒ CM2 = $\frac{125 - 45}{4}$ ⇒ CM2 = 20 ⇒ CM = $2 \sqrt{5}$ cm

In ÃÂÃÂÃÂÃÂÃÂÃÂABC ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 90ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and AD ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ¥ BC where D lies on BC. If BC = 8 cm, AD = 6 cm, then arÃÂÃÂÃÂÃÂÃÂÃÂABC : arÃÂÃÂÃÂÃÂÃÂÃÂACD = ?

(A) 4 : 3
(B) 25 : 16
(C) 16 : 9
(D) 25 : 9

Solution: According to question, ΔABC ∼ ΔACD $\begin{aligned} ∴ \frac{area of Δ A B C}{area of Δ A C D} = \frac{B C^{2}}{A D^{2}} \\ \frac{area of Δ A B C}{area of Δ A C D} = \frac{8^{2}}{6^{2}} \\ \frac{area of Δ A B C}{area of Δ A C D} = \frac{64}{36} \\ \frac{area of Δ A B C}{area of Δ A C D} = \frac{16}{9} \end{aligned}$ area ΔABC : area ΔACD = 16 : 9

In a triangle ABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ A = 70ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ B = 80ÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and D is the incenter of ÃÂÃÂÃÂÃÂÃÂÃÂABC, ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ ACB = 2xÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ° and ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ BDC = yÃÂÃÂÃÂÃÂÃÂÃÂÃÂÃÂ°. The values of x and y, respectively are:

(A) 15°, 130°
(B) 15°, 125°
(C) 35°, 40°
(D) 30°, 150°

Solution: ∠C = 180 - (∠A + ∠B) ∠C = 180 - 150 2x = 30 x = 15° ∠BDC = 90° + $\frac{1}{2}$ ∠A ∠BDC = 90° + $\frac{1}{2}$ × 70° ∠BDC = 90° + 35° ∠BDC = 125° So value of x and y are = 15°, 125°

In ÃÂÃÂÃÂÃÂÃÂÃÂABC and ÃÂÃÂÃÂÃÂÃÂÃÂDEF, AB = DE and BC = EF, then one can infer that ÃÂÃÂÃÂÃÂÃÂÃÂABC ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂ ÃÂÃÂÃÂÃÂÃÂÃÂDEF, when

(A) ∠BAC = ∠EFD
(B) ∠ACB = ∠EDF
(C) ∠ABC = 2∠DEF
(D) ∠ABC = ∠DEF

ABC is a right angled triangled, right angled at C and P is the length of the perpendicular from C on AB. If a, b and c are the length of the sides BC, CA and AB respectively, then

(A) $\frac{1}{p^{2}} = \frac{1}{b^{2}} - \frac{1}{a^{2}}$
(B) $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
(C) $\frac{1}{p^{2}} + \frac{1}{a^{2}} = - \frac{1}{b^{2}}$
(D) $\frac{1}{p^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}$

Solution: According to question, ACB is a right angle triangle ∴ area of ΔACB $\frac{1}{2}$ × AC × BC = $\frac{1}{2}$ × AB × PC $\frac{1}{2}$ × b × a = $\frac{1}{2}$ × c × p c = $\frac{a b}{p}$ . . . . . . . (i) By using pythagoras theorem AB2 = AC2 + BC2 c2 = b2 + a2 . . . . . . . . . . (ii) Put the value of C in equation (ii) $\begin{aligned} {(\frac{a b}{p})}^{2} = a^{2} + b^{2} \\ \Rightarrow \frac{a^{2} b^{2}}{p^{2}} = a^{2} + b^{2} \\ \Rightarrow \frac{1}{p^{2}} = \frac{a^{2}}{a^{2} b^{2}} + \frac{b^{2}}{a^{2} b^{2}} \\ \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \end{aligned}$ Alternate : From figure, $\begin{aligned} P = \frac{a b}{c} \\ P = \frac{a b}{\sqrt{a^{2} + b^{2}}} (∵ a^{2} + b^{2} = c^{2}) \\ P^{2} = \frac{a^{2} b^{2}}{a^{2} + b^{2}} \\ \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} \end{aligned}$

If the circumcentre of a triangle lies outside it, then the triangle is

(A) Equilateral
(B) Acute angled
(C) Right angled
(D) Obtuse angled

The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ÃÂÃÂÃÂÃÂ¢ÃÂÃÂÃÂÃÂÃÂÃÂ PSR is

(A) 30°
(B) 15°
(C) 60°
(D) 45°

Solution: According to question, Given : PQR is an equilateral triangle QR = RS PR = RS ∠SRP = 180° - 60° (Exterior ∠) ∠SRP = 120° ∴ ∠RPS = ∠RSP ∴ ∠RPS + ∠PRS + ∠RSP = 180° 2∠PSR = 180° - 120° ∠PSR = $\frac{60^{\circ}}{2}$ ∠PSR = 30°

10.

If in a triangle, the orthocentre lies on vertex, then the triangle is

(A) Acute angled
(B) Isosceles
(C) Right angled
(D) Equilateral

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Practice MCQ Questions and Answer on Triangles

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