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Q1. An isosceles triangle with base 48 cm has area 240 sq cm. Find the remaining two sides of the triangle.

Solution

Q2. Two sides of a triangle are 8 cm and 11 cm and its perimeter is 32 cm. Find the area of the triangle.

Solution

Perimeter = 32 cm, a = 8 cm, b = 11 cm = 16 cm Area of Area of triangle = begin mathsize 12px style 8 square root of 30 space cm squared end style
Q3. Find the area of a triangle whose sides are 8 cm, 11 cm, 13 cm. Hence, find the altitude using longest side as base.

Solution

Using Heron's formula, Height = 6.74 cm
Q4. A triangular park has sides 120 m, 80 m, 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per meter leaving a space 3 m wide for a gate on one side.

Solution

Here, Therefore, area of park Perimeter of park = 50 + 80 + 120 = 250 m Thus, length of wire needed = 250 - 3 = 247 m So, cost of fencing = Rs 20 247 = Rs 4940
Q5. Find the area of quadrilateral right angled at B, in which AB = 5 cm, BC = 12 cm, AD = 22 cm and CD = 13 cm.

Solution

Q6. The area of an isosceles triangle with base 10 cm and perimeter 36 cm is:
  • 1) 360 sq cm
  • 2) 65 sq cm
  • 3) 60 sq cm
  • 4)

Solution

straight s equals 36 over 2 equals 18 comma space and space the space sides space are space 10 comma space 13 comma space 13 space cm. By space Heron apostrophe straight s space formula comma space Area space equals square root of 18 cross times 8 cross times 5 cross times 5 end root space equals space 5 cross times 3 cross times 4 space equals space 60 space sq space cm.
Q7. Find the percentage increase in the area of triangle if its each side is doubled.

Solution

Q8. Find the area of an equilateral triangle with side 10 cm.

Solution

Side = 10 cm S = Area of triangle = =
Q9. Find the area of a triangle whose sides are 6.5 cm, 7 cm and 7.5 cm.

Solution

Here, P = 6.5 + 7 + 7.5 = 21 cm begin mathsize 12px style straight S equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction end style So, s = begin mathsize 12px style 21 over 2 end style= 10.5 cm s - a = 4, s - b = 3.5, s - c = 3 Area of triangle = begin mathsize 12px style square root of straight S open parentheses straight S minus straight a close parentheses open parentheses straight S minus straight b close parentheses open parentheses straight S minus straight c close parentheses end root end style  Î”= 21 cm2
Q10. Calculate the area of the rhombus of sides 41 cm and length of shorter diagonal is 18 cm.

Solution

Q11. The parallel sides of an isosceles trapezium are 16 and 28 units and the equal side is 24 units. Find the height of the trapezium.

Solution

Q12. Heron's formula is:
  • 1)
  • 2)
  • 3)
  • 4)

Solution

Q13. The sides of a triangle are in the ratio 3:5:7 and it's perimeter is 300m. Find its area.

Solution

Ratio of sides = 3:5:7 Let sides be 3x, 5x, 7x So 3x + 5x + 7x = 300m 15x = 300m x = 20 m Sides are 60m, 100m, 140m s =
Q14. The sides of a triangular ground are 5m, 7m and 8 m respectively. Find the cost of leveling the ground at the rate of Rs.10 per m2. (use = 1.73).

Solution

Cost of leveling the ground = 17.3 10 = Rs 173
Q15. A triangular park in a city has dimensions 30 m x 26 m x 28 m. A gardener has to plant grass inside the park at Rs. 1.50 per m2. Find the amount to be paid to the gardener.

Solution

Here lengths of the sides of park are 30m, 26m, 28m. begin mathsize 12px style straight S equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction end style Therefore, area of the triangular park The amount to be paid to the gardener= Rs.
Q16. The unequal side of an isosceles is 6 cm and its perimeter is 24 cm. Find the area.

Solution

Since, P = 24 cm So, sum of equal sides = 24 - 6 = 18 cm Length of equal side = 9 cm Thus sides are: 9 cm, 9 cm and 6 cm. Also, s = 12 cm
Q17. Find the area of a triangle with perimeter 22 cm, one side 9 cm and difference of other two sides is 7 cm.

Solution

Let a = 9 Now, a + b + c = 22 b + c = 22 - 9 = 13 b + c = 13           …(1) Also, b - c = 7(given)… (2) Adding (1) and (2) we get 2b = 20 Or, b = 10 From (1), c = 13 - 10 = 3. Hence, a = 9, b = 10, c = 3 and s = 11  
Q18. Find the area of a triangle, two sides of which are 60 cm and 100 cm and the perimeter is 300 cm.

Solution

a + b + c =300 Third side = 300 - 60 - 100= 140cm Area = = = = = 1500 cm2
Q19. A garden is in the shape of quadrilateral. The sides of the garden are 9m, 40m, 28m and 15m respectively in consecutive order and the angle between first two sides is a right angle. Find the area of the garden.

Solution

Let ABCD be the garden. Area of the garden = 180 + 126 = 306 m2
Q20. The base of an isosceles triangle measures 24 cm and its area is 60 cm2. Find its perimeter.

Solution

Let each equal side of isosceles begin mathsize 12px style straight S equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction end style So, Semi-perimeter = (x+12) cm= (13+12) cm= 25 cm Perimeter = 2 x 25 cm = 50 cm.
Q21. Find the height of the trapezium in which parallel sides are 25 cm and 10 cm and non-parallel sides are 14 cm and 13 cm.

Solution

Draw AE||BC, through A. Draw AF CD For ADE, s = 21 cm, Area of ADE= = = 84 cm2 base height = 84 cm2 Height = = 11.2 cm
Q22. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Solution

Perimeter of a triangle = 30 cm 12+12+Third side = 30
Q23. Sides of a triangle are in the ratio 13 : 14 : 15 and its perimeter is 84 cm. Find its area.

Solution

Here, perimeter of a triangle = 84 cm Sides are in the ratio 13:14:15 Sum of ratio = 42 cm By Heron's formula Area = Hence, area of a triangle is 336.
Q24.

Solution

Cost of one square cm paper = Rs 5 Hence, cost of 117.6 sq cm paper = Rs 5 x 117.6 = Rs 588
Q25. The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm and the diagonal AC measure 42 cm. Find the area of the parallelogram.

Solution

Let a= 20 cm b= 34 cm c=42 cm Diagonal of gm divides it into two congruent triangles
Q26. A floral design on the ceiling is made of 6 isosceles triangles. Each triangle has perimeter 100cm and base 48cm. If the triangles are painted green and blue alternately, then find the area painted green.

Solution

Q27. The sides of a triangle are in the ratio 4:5:5 and its perimeter is 168 m. Find the area of the triangle.

Solution

Length of sides are: begin mathsize 12px style 4 over 14 cross times 168 equals 48 space straight m comma space 5 over 14 cross times 168 equals 60 space straight m comma space fraction numerator begin display style 5 end style over denominator begin display style 14 end style end fraction cross times 168 equals 60 space straight m end style Therefore, a = 48 m, b = 60 m and c = 60 m begin mathsize 12px style straight S equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction end style S = begin mathsize 12px style fraction numerator 48 plus 60 plus 60 over denominator 2 end fraction equals 84 end style Area of =
Q28. A popular mathematics club decided to gift each member with a flag. The flag is of the shape of an isosceles triangle and looks like:

Solution

Q29. Find the height of a trapezium with parallel sides 10 and 18 units, adjacent sides 7 and 9 units. Also find the area of the trapezium.

Solution

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