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Heron’s Formula | Heron’s Formula - Exercise 12.2

Question 7

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm

and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much

paper of each shade has been used in it?

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As the kite is in the shape of a square, its area will be

\begin{array}{l} A=(1 / 2) \times(\text { diagonal })^{2} \\ \Rightarrow \text { Area of the kite }=(3 / 2) \times 32 \times 32=512 \mathrm{cm}^{2} \end{array}

The area of shade I = Area of shade II

\begin{aligned} &\Rightarrow 512 / 2 \mathrm{cm}^{2}=256 \mathrm{cm}^{2}\\ &\text { So, the total area of the paper that is required in each shade }=256 \mathrm{cm}^{2} \end{aligned}

For the triangle section (III),

The sides are given as 6 cm, 6 cm and 8 cm

Now, the semi perimeter of this isosceles triangle = (6 + 6 + 8)/2 cm = 10 cm

By using Heron's formula, the area of the III triangular piece will be

= √[s (s-a) (s-b) (s-c)]

\begin{array}{l} =\sqrt{10}(10-6)(10-6)(10-8) \mathrm{cm}^{2} \\ =\sqrt{10 \times 4 \times 4 \times 2 \mathrm{cm}^{2}} \\ =8 \mathrm{v} 6 \mathrm{cm}^{2} \end{array}

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Our top 5% students will be awarded a special scholarship to Lido.

subject-cta

Question 7

A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm

and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much

paper of each shade has been used in it?

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

As the kite is in the shape of a square, its area will be

\begin{array}{l} A=(1 / 2) \times(\text { diagonal })^{2} \\ \Rightarrow \text { Area of the kite }=(3 / 2) \times 32 \times 32=512 \mathrm{cm}^{2} \end{array}

The area of shade I = Area of shade II

\begin{aligned} &\Rightarrow 512 / 2 \mathrm{cm}^{2}=256 \mathrm{cm}^{2}\\ &\text { So, the total area of the paper that is required in each shade }=256 \mathrm{cm}^{2} \end{aligned}

For the triangle section (III),

The sides are given as 6 cm, 6 cm and 8 cm

Now, the semi perimeter of this isosceles triangle = (6 + 6 + 8)/2 cm = 10 cm

By using Heron's formula, the area of the III triangular piece will be

= √[s (s-a) (s-b) (s-c)]

\begin{array}{l} =\sqrt{10}(10-6)(10-6)(10-8) \mathrm{cm}^{2} \\ =\sqrt{10 \times 4 \times 4 \times 2 \mathrm{cm}^{2}} \\ =8 \mathrm{v} 6 \mathrm{cm}^{2} \end{array}

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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