Question

ABC Is an isosceles triangle having AB = BC = 10 cm ∠ABC = 30 °. Calculate the area of ΔABC.
​

Answer 1

Answer:

Let in triangle ABC , AB = AC = x (let) , D is the mid point of BC.

In triangle CDA

CD/AC=sin15°

5/AC= sin(45°-30°)=(3^1/2 - 1)/2.2^1/2

AC = x =(10.2^1/2)/(3^1/2 -1)

x=(10.2^1/2) (3^1/2+1)/(3^1/2 -1) (3^1/2 +1)

x=(10.2^1/2) (3^1/2 +1)/2 =5.2^1/2(3^1/2+1)

Area of triangle ABC =(1/2).AB.AC.sin30°

= (1/2).x.x.(1/2)

= (1/4).x^2 =(1/4).50.(3^1/2 + 1)^2 cm^2

=(50/4)(3+1+2.3^1/2)

=(25/2).2.(2+3^1/2)

=25.(2+3^1/2) cm^2.

Answer 2

$\dashrightarrow$ Steps :-

• First draw a line segment AB of 10 cm.
• Next, take a protractor and keep it's mid-point on the point B and draw a ray measuring 30°.
• Then, take a compass and measure it as 10 cm and cut an arc on the ray drawn.
• Then, join the point A where the ray and the arc meets. The point where the arc and the ray meets is named as point C.
• Finally, the triangle is constructed.

$\dashrightarrow$ Instruments used :-

• Ruler :- Ruler is used to draw the line segments of accurate measurements. The lines should also be straight.
• Protractor :- Protractor is used to draw the ray of accurate degrees or measurements. It is also used to measure the angles.
• Compàss :- Compàss is used to draw the arc on the ray by taking measurement. Here, it's used to draw the arc of 10 cm on the ray.

Area of the triangle :-

CD/BC = sin30°

CD/BC = (1/2)

First, find the height of the triangle

CD = ½ $\times$ BC

CD = ½ $\times$ 10

CD = 5 cm

${\tt \longrightarrow \dfrac{1}{2} \times (base \times height)}$

${\tt \longrightarrow \dfrac{1}{2} \times (10 \times 5)}$

${\tt \longrightarrow \dfrac{1}{2} \times 50 = \boxed{\tt 25 \: {cm}^{2}}}$

$\Huge\therefore$ The area of the triangle is 25 cm².