WAEC Mathematics Question 2015 ( Essay Question Five 5 )

May 23, 2020

By Kings

WAEC Mathematics 2015, WAEC Mathematics 2015 Essay

Question 5

A trapezium PQRS is such that PQ // RS and the perpendicular P to RS is 40 cm. If |PQ| = 20 cm, |SP| = 50cm and |SR| = 60 cm. Calculate, correct to 2 significant figures, the;

(a) Area of the trapezium ;

(b) < QRS.

SOLUTION

By interpretation, the diagram of the trapezium is represented as

(a)

Area of the trapezium is given by;

$A=\frac{1}{2}(a+b)h$

Where a= 20cm, b = 60cm, h = 40cm

Substituting the values, we have;

$A=\frac{1}{2}(20+60)40$

$A = \frac{1}{2}(80)40$

$A = \frac{80\times40}{2}$

$A = 80\times20$

A = 1600 $cm^{2}$ ANSWER

(b)

To obtain $\angle QRS$

Note that PQ = MN = 20cm

solving for |SM|

Considering $\bigtriangleup SMP$ , we solve using pythagoras theorem:

$|PS|^{2} = |SM|^{2}+|PM|^{2}$

$50^{2} = |SM|^{2}+40^{2}$

$|SM|^{2} = 50^{2} - 40^{2}$

$|SM|^{2} = 2500 - 1600$

$|SM|^{2} = 900$

$|SM| = \sqrt{900}$

$|SM| = 30cm$

But SR = 60cm = |SM| + |MN| + |NR|

$\therefore$ 60 = 30 + 20 + |NR|

60 = 50 + |NR|

|NR| = 60 - 50

|NR|= 10cm

Solving for |QR|

We consider $\bigtriangleup QNR$ and solve using pythagoras theorem;

$|QR|^{2}=|QN|^{2}+|NR|^{2}$

$|QR|^{2}=40^{2}+10^{2}$

$|QR|^{2}$ = 1600 + 100

$|QR|^{2}$ = 1700

$|QR|=\sqrt{1700}$

|QR| = 41.23cm

To solve for |QS|, We consider $\bigtriangleup QNS$ and solve using pythagoras theorem;

$|QS|^{2}=|SN|^{2}+|QN|^{2}$

But |SN| + |SM|+|MN|

|SN| = 30+20=50cm

$\therefore |QS|^{2}=50^{2}+40^{2}$

$|QS|^{2}=2500+1600$

$|QS|^{2}=4100$

$|QS|=\sqrt{4100}$

$|QS|=64.03cm$

Hence, to solve for $\ominus$ in $\bigtriangleup QRS$

We solve using the COSINE RULE

$r^{2}=s^{2}+q^{2}-2sqCosR^{o}$

$64.03^{2}=41.23^{2}+60^{2}-2(41.23\times 60)Cos\ominus ^{o}$

4099.84 =1699.91+3600-4947.6Cos $\ominus$

4099.84 =5299.91-4947.6Cos $\ominus$

4099.84-5299.91=-4947.6Cos $\ominus$

-1200.07=-4947.6Cos $\ominus$

1200.07 = 4947.6Cos $\ominus$

$\therefore$ Cos $\ominus$ = $\frac{1200.07}{4947.6}$

Cos $\ominus$ = 0.2426

$\ominus$ = $\cos^{-1}(0.2426)$

$\ominus$ = 75.96

$\ominus \simeq 76^{o}$ ANSWER

Hence, $\angle QRS$ = $\ominus \simeq 76^{o}$

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