Solved Examples: Triangles

Question 1: In a $△$ ABC, D and E are two points on sides AB and AC such that DE is parallel to BC and AD : DB = 2 : 1. If AE = 8 cm, then find the length of AC.

a) 12 cm
b) 10 cm
c) 16 cm
d) 20 cm

Answer (A)

Solution.

The line DE is parallel to BC, so triangle ADE is similar to triangle ABC by the Basic Proportionality (Thales) theorem.

\(\displaystyle \frac{AD}{AB}=\frac{AE}{AC}\)

Given \(AD:DB=2:1\). Hence \(AD:AB=2:(2+1)=2:3\).

\(\displaystyle \frac{AE}{AC}=\frac{2}{3}\)

AE = 8 cm, so

\(\displaystyle AC=\frac{3}{2}\times AE=\frac{3}{2}\times 8=12\ \text{cm}\)

Question 2: In a $△$ ABC, Points D and E are on sides AB and AC such that DE is parallel to BC and AD : DB = 3 : 1 and AE = 18 cm. Then find AC.

a) 26 cm
b) 24 cm
c) 28 cm
d) 32 cm

Answer (B)

Solution.

DE ∥ BC ⇒ △ADE ∼ △ABC.

\(\displaystyle \frac{AD}{AB}=\frac{AE}{AC}\)

Given \(AD:DB=3:1\). Hence \(AD:AB=3:(3+1)=3:4\).

\(\displaystyle \frac{AE}{AC}=\frac{3}{4}\)

AE = 18 cm, so

\(\displaystyle AC=\frac{4}{3}\times AE=\frac{4}{3}\times 18=24\ \text{cm}\)

Question 3: In a $△$ ABC, points D and E are on the sides of AB and AC respectively such that DE is parallel to BC and AD : AB = 2 : 5 and AE = 4 cm. Then find AC.

a) 10 cm
b) 14 cm
c) 12 cm
d) 9 cm

Answer (A)

Solution.

DE ∥ BC ⇒ △ADE ∼ △ABC.

\(\displaystyle \frac{AD}{AB}=\frac{AE}{AC}\)

Given \(AD:AB=2:5\). Hence \(\displaystyle \frac{AD}{AB}=\frac{2}{5}\).

\(\displaystyle \frac{AE}{AC}=\frac{2}{5}\)

AE = 4 cm, so

\(\displaystyle AC=\frac{5}{2}\times AE=\frac{5}{2}\times 4=10\ \text{cm}\)

Question 4: The coordinates of the vertices of a right-angled triangle are A (6, 2), B(8, 0) and C (2, -2). The coordinates of the orthocentre of triangle PQR are

a) (2, -2)
b) (2, 1)
c) (6, 2)
d) (8, 0)

Answer (C)

Solution.

For a triangle with vertices A(6,2), B(8,0), C(2,-2), check which angle is right-angled by using dot product of vectors from a vertex.

Vector AB = (8-6,\;0-2) = (2,\;-2)

Vector AC = (2-6,\;-2-2) = (-4,\;-4)

Dot product AB·AC = \(2\times(-4)+(-2)\times(-4)= -8+8=0\)

Since the dot product is 0, ∠A is a right angle.

In a right-angled triangle the orthocentre is the vertex at the right angle.

Therefore orthocentre = A = (6, 2).

Question 5: find the area of an equilateral triangle if the height of the triangle is 24 cm.
a)

b)

c)

d)

Answer (A)

Solution.

For an equilateral triangle of side \(a\), height \(h=\dfrac{\sqrt{3}}{2}a\).

\(\displaystyle a=\frac{2h}{\sqrt{3}}\)

Given \(h=24\ \text{cm}\), so

\(\displaystyle a=\frac{2\times 24}{\sqrt{3}}=\frac{48}{\sqrt{3}}=16\sqrt{3}\ \text{cm}\)

Area \(=\dfrac{\sqrt{3}}{4}a^2\).

\(\displaystyle \text{Area}=\frac{\sqrt{3}}{4}\times(16\sqrt{3})^{2}=\frac{\sqrt{3}}{4}\times(256\times 3)=\frac{\sqrt{3}}{4}\times 768=192\sqrt{3}\ \text{cm}^2\)

Question 6: Three sides of a triangular meadow are of length 28 m, 45 m and 53 m long respectively. Find the cost of sowing seeds(in rupees per sq.m) in the meadow at the rate of 12 rupees per sq.m.

a) 7560
b) 6860
c) 7960
d) 7860

Answer (A)

Solution.

Check whether the triangle is right-angled. For sides 28, 45, 53:

\(\displaystyle 28^{2}+45^{2}=784+2025=2809\)

\(\displaystyle 53^{2}=2809\)

Since \(28^{2}+45^{2}=53^{2}\), the triangle is right-angled with legs 28 m and 45 m.

Area \(=\dfrac{1}{2}\times(\text{product of legs})=\dfrac{1}{2}\times 28\times 45\)

\(\displaystyle \text{Area}=14\times 45=630\ \text{m}^2\)

Cost at ₹12 per m^2 is

\(\displaystyle \text{Cost}=630\times 12=7560\ \text{rupees}\)

Question 7: In a triangle PQR, internal angular bisectors of $\angle 𝑄$ and $\angle 𝑅$ intersect at a point O. If $\angle 𝑃$=$110\circ$ then what is the value of $\angle 𝑄𝑂𝑅$ ?
a) $125<sup\; style="font-size:13px\; !important">\circ </sup>$
b) $135<sup\; style="font-size:13px\; !important">\circ </sup>$
c) $145<sup\; style="font-size:13px\; !important">\circ </sup>$
d) $115<sup\; style="font-size:13px\; !important">\circ </sup>$

Answer (C)

Solution.

O is the incenter (intersection of internal bisectors of Q and R).

For an incenter, the angle formed by the intersection of the bisectors at O opposite vertex P is

\(\displaystyle \angle QOR=90^\circ+\frac{1}{2}\angle P\)

Given \(\angle P=110^\circ\).

\(\displaystyle \angle QOR=90^\circ+\frac{110^\circ}{2}=90^\circ+55^\circ=145^\circ\)

Question 8: In a triangle XYZ, XA is the angle bisector onto YZ. If the semiperimeter of the triangle is 12 and XY=12 ,YZ=6 then what is the ratio of YA:AZ ?

a) 2:3
b) 2:1
c) 1:2
d) 3:2

Answer (B)

Solution: Semiperimeter \(s=12\). So \(x+y+z=24\) where \(x=XY,\;y=YZ,\;z=XZ\).

Given \(XY=12,\;YZ=6\).

\(\displaystyle 12+6+XZ=24\)

\(\displaystyle XZ=6\)

By the Angle Bisector Theorem,

\(\displaystyle \frac{YA}{AZ}=\frac{XY}{XZ}\)

\(\displaystyle \frac{YA}{AZ}=\frac{12}{6}=2:1\)

Question 9: In a triangle ABC, AX is the angle bisector onto BC. If the semiperimeter of the triangle is 9 and AB=4 ,BC=6 then what is the ratio of BX:XC ?

a) 2:3
b) 2:1
c) 1:2
d) 3:2

Answer (C)

Solution.

Semiperimeter \(s=9\). So \(a+b+c=18\) where \(AB=4,\;BC=6,\;AC=\ ?\)

\(\displaystyle 4+6+AC=18\)

\(\displaystyle AC=8\)

By the Angle Bisector Theorem,

\(\displaystyle \frac{BX}{XC}=\frac{AB}{AC}=\frac{4}{8}=\frac{1}{2}\)

Hence \(BX:XC=1:2\).

Question 10: In an equilateral triangle,if h-R=15 cm where h=height of the triangle and R=circumradius then what is the area of the triangle ?

a) 

b) 

c) 

d) 

Answer (D)

Solution.

For an equilateral triangle of side \(a\),

\(\displaystyle h=\frac{\sqrt{3}}{2}a\)

\(\displaystyle R=\frac{a}{\sqrt{3}}\)

So

\(\displaystyle h-R=\frac{\sqrt{3}}{2}a-\frac{a}{\sqrt{3}}=a\left(\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}}\right)\)

Simplify the coefficient:

\(\displaystyle \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{3}}=\frac{3-2}{2\sqrt{3}}=\frac{1}{2\sqrt{3}}\)

\(\displaystyle h-R=\frac{a}{2\sqrt{3}}\)

Given \(h-R=15\ \text{cm}\), so

\(\displaystyle a=30\sqrt{3}\ \text{cm}\)

Area \(=\dfrac{\sqrt{3}}{4}a^{2}\)

\(\displaystyle \text{Area}=\frac{\sqrt{3}}{4}\times(30\sqrt{3})^{2}=\frac{\sqrt{3}}{4}\times(900\times 3)=\frac{\sqrt{3}}{4}\times 2700=675\sqrt{3}\ \text{cm}^2\)