4 Marks Questions

Question 14 Marks

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours each piece measuring 20cm, 50cm and 50cm. How much cloth of each colour is required for the umbrella?

Answer

Semi perimeter of each triangular piece $=\frac{(50+50+20)}{2\text{cm}}=\frac{120}{2\text{cm}}=60\text{cm}$
Using heron's formula,
Area of the triangular piece $=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{60(60-50)(60-50)(60-20)\text{cm}^2}$
$=\sqrt{60\times10\times10\times40\text{cm}^2}$
$=200\sqrt{6\text{cm}^2}$
Area of triangular piece $=5\times200\sqrt{6\text{cm}^2}=1000\sqrt{6\text{cm}^2}$

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Question 24 Marks

An isosceles triangle has perimeter 30cm and each of the equal sides is 12cm. Find the area of the triangle.

Answer

Let the third side of this triangle be x.
Perimeter of triangle = 30cm
12cm + 12cm + x = 30cm
x = 6cm
$\text{s}=\frac{\text{Perimeter of triangle}}{2}=\frac{30\text{cm}}{2}=15\text{cm}$
By Heron's formula,
$\text{Area of triangle}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\big[\sqrt{15(15-12)(15-12)(15-6)}\big]\text{cm}^2$
$=\big[\sqrt{15(3)(3)(9)}\big]\text{cm}^2$
$=9\sqrt{15}\text{cm}^2$

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Question 34 Marks

Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer

Let the third side of the triangle be x.
Perimeter of the given triangle = 42cm
18cm + 10cm + x = 42
x = 14cm
$\text{s}=\frac{\text{Perimeter}}{2}=\frac{42\text{cm}}{2}=21\text{cm}$
By Heron,s formula,
$\text{Area of triangle}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$\text{Area of the given triangle}=\big(\sqrt{21(21-18)(21-10)(21-14)}\big)\text{cm}^2$
$=\big(\sqrt{21(3)(11)(7)}\big)\text{cm}^2$
$=21\sqrt{11}\text{cm}^2$

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Maths 4 Marks Questions

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