CIE Mathematics Paper-1: Specimen Questions 159 - 160 of 175

Question number: 159

Essay Question▾

Describe in Detail

Solve:

A trapezoid made by different length lines

A trapezoid made by different length lines

Find the area and perimeter of given trapezoidal

Solve:

(a) The perimeter,

(b) The area.

Explanation

Here,

(a) A perimeter can find by adding the length of all shapes or boundaries.

So, here, perimeter = AB+BC+CD+DA

Where, AB = 18 cm, BC = 14 cm, CD = 12 cm, DA = 14 cm.

So, Perimeter = 18+14+12+14=58cm.

(b) Area of Trapezoid ABCD = a+b2×h , where, a and b = two parallel sides, h = height.

So, here, a = 12 cm, b = 18 cm and h = 12 cm.

So, Area A = 12+182×12=302×12=15×12=180cm.

Question number: 160

Essay Question▾

Describe in Detail

Solve

A geometrical shape made by lines

A geometrical shape made by lines

Find the area and perimeter of given geometrical shape

Find out:

(a) The perimeter,

(b) The area.

Explanation

Here,

A geometrical shape made by different length lines

A geometrical shape made by different length lines

Find the area and perimeter of given geometrical shape

(a) A perimeter can find by adding the length of all shapes or boundaries.

So, here, perimeter = AB+BC+CD+DE+EF+FA

Where, AB = 10 cm, BC = 22 cm, CD = 10 – 6 = 4 cm, DE = 22 – 14 = 8 cm, EF = 6 cm, FA = 14 cm.

So, Perimeter = 10+22+4+8+6+14=64cm.

(b) Area of ABCDEF = Area of ABGF (rectangle) + Area of EGCD (rectangle)

So, area of ABGF = AB×BGorAF×FG=10×14=140cm2

Now, area of EGCD = EG×GCorED×DC=8×4=32cm2

So, AreaofABCDEF=140+32=172cm2 .

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