# Question-1

**Find the area of a triangle in which**

(i) a = b = c = 4 cm

(ii) In right angled triangle, hypotenuse is 13 cm and one side is 5 cm.

(i) a = b = c = 4 cm

(ii) In right angled triangle, hypotenuse is 13 cm and one side is 5 cm.

**Solution:**

(i) Area of a equilateral triangle = = Â´ (4)

^{2 }= 6.928 cm

^{2}

(ii) Area of right angled triangle = Â´ base Ã— height

Hypotenuse = 13cm, one side = 5cm

(Hypotenuse)

^{2}= (one side)

^{2 }+ (other side)

^{2}

13

^{2 }= x

^{2 }+ 5

^{2}

⇒ x = 12 cm

Area of the right angled triangle= Â´ 12 Â´ 5 = 30 cm

^{2}.

# Question-2

**The cost of levelling a park at the rate of**`**2/km**`^{2}is**2,700. If the park is in right angled triangular form with one side being 45 km, find the hypotenuse.****Solution:**

The cost of leveling = `2,700

The rate of leveling = `2/km

^{2}

Area of the park = Cost/Rate = 2700/2 = 1350 km

^{2 }

One side of the park = 45 km

Let the other side = x km

Area of the right angled

**triangular park = Ã— one side Ã— other side**

.

^{.}. 1350 = Ã— 45 Ã— x â‡’ x = 60 km

Again (Hypotenuse)

^{2 }= (I

^{st}side)

^{2}+(IInd side)

^{2}

= 45

^{2 }+ 60

^{2 }= 5625

.

^{.}. Hypotenuse = 75 km

# Question-3

**The perimeter of a triangular field is 240 m. It's two sides are 78 m and 50 m. Find the length of the altitude on the side of 50 m length from its opposite vertex.**

**Solution:**

Let the three sides of the triangle be a, b, c, then perimeter = a + b + c.

.

^{.}. 240 = 50 + 78 + c or c = 112 m

Area of the triangle =

Or Â´ base Â´ height =

Ãž Â´ 50 Â´ h =

âˆ´ h = 67.2 m

# Question-4

**(i) The area of an equilateral triangle is 100m**

(ii) Find the base of an isosceles triangle whose area is 60 sq.cm and length of equal sides is 13 cm.

(iii) The area of an isosceles right triangle is 200 m

(iv) The ratio of the base to the altitude of a triangular field is 3 : 1. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10. Find its base and height.^{2}. Find the perimeter of the triangle.(ii) Find the base of an isosceles triangle whose area is 60 sq.cm and length of equal sides is 13 cm.

(iii) The area of an isosceles right triangle is 200 m

^{2}. Find its hypotenuse.(iv) The ratio of the base to the altitude of a triangular field is 3 : 1. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10. Find its base and height.

**Solution:**

(i) Area of the equilateral triangle = (side)

^{2}.

100 = (a)

^{2}

Ãž a = 20 m

âˆ´ Perimeter of the equilateral triangle = 3a = 3 Â´ 20 = 60 m

(ii) We are given an isosceles Î”ABC in which

AB = AC = 13 cm.

Let AD ^ BC and BC = 2x

â‡’ BD = DC = x cm

AD =

Area of Î”ABC = Â´ base Â´ height

â‡’ 60 =

â‡’ x = 12 or 5

.

^{.}. base = 24 cm or 10 cm

(iii)

Let AB=BC=x cm

AC

^{2 }= AB

^{2 }+ BC

^{2}

= x

^{2 }+ x

^{2}

â‡’ AC = âˆš2x

Area of the triangle = Â´ base Â´ height

Or 200 = Ã— x Ã— x â‡’ x = 20

.

^{.}. Hypotenuse âˆš2x = âˆš2 Ã—20 = 28.28 m

(iv) Let the altitude be = x m

Then the base of the triangle = 3x m

Area of the field =

= 13.5 hectare

= 135000 m

^{2}

Ãž Â´ base Ã— height = 135000

Ãž Â´ 3x Ã— x = 135000

â‡’ x = 300

âˆ´ Base(3x) = 900 m

Altitude (x) = 300 m

# Question-5

**(i) The cross section of a canal is in a trapezium shape. If the canal is 10 m wide at the top, 6 m wide at the bottom and area of cross section is 72 sq. m, find the depth of the canal.**

(ii) Find the area of a trapezium whose parallel sides are 12 cm and 16 cm and the distance between the parallel sides in 10 cm.(ii) Find the area of a trapezium whose parallel sides are 12 cm and 16 cm and the distance between the parallel sides in 10 cm.

**Solution:**

(i) Area of a trapezium = (sum of parallel sides) Ã— height

or 72 = (6 + 10) Â´ h

Ãž h = 9 m

(ii) Area of trapezium = (a + b)Ã—h

= (12 + 16)Ã—10

= 140 cm

^{2 }

# Question-6

**The area of a quadrilateral is 360 m**^{2}and the perpendiculars drawn to one of the diagonal from the opposite vertices are 10 m and 8 m. Find the length of the diagonal.**Solution:**

In the given Fig. AE = h

_{1 }= 8 cm

CF = h

_{2 }= 10 cm and area of quadrilateral ABCD = 360 m

^{2}

Area of a quadrilateral = (Diagonal)(h

_{1 }+ h

_{2})

360 = Â´ DB Â´ (10 + 8)

â‡’DB = 40 m

# Question-7

**The area of a trapezium is 1700 m**^{2}. If one of the parallel sides is 64 m and the distance between the parallel sides is 34 m, find the length of the other parallel side.**Solution:**

ABCD is a trapezium in which DC = 64 m and altitude

AE = 34 m

Let other parallel side AB = x m

Area (ABCD) = (AB + DC) Ã— AE

1700 = (x + 64)Ã—34

or x = 36 m.

# Question-8

**The parallel sides of a trapezium is 77 m and 60 m and its non-parallel sides are 26 m and 25 m. Find the area of the trapezium.****Solution:**

Let ABCD be a trapezium in which

AB = 60 m, CD = 77 m, BC = 26 m, AD = 25 m

and

Draw BE || to AD

Thus ABED is parallelogram

Hence AB = DE = 60 m.

AD = BE = 25 m

and EC = DC - DE = 77 - 60 = 17 m

and

Now in Î”BEC,

BE = 25 m, BC = 26m, EC = 17m

Let the height of B from EC be = h m

S = = 34

Area of Î”BEC =

Â´ EC Â´ h =

⇒ Â´ 17 Â´ h =

or h = 24 m.

Now area of trapezium = (sum of parallel sides) Ã— height

\ Area of trapezium = (60 + 77) Â´ 24 = 1644 m

^{2}.