mensuration Model Questions & Answers, Practice Test for ssc chsl tier 1 2024

Answer: (d)

∠BOD = 180° – 106° = 74°

Since, ∠BOD is an angle made by arc BD on centre.

Here, ∠BCD is an angle made by arc BD on circumference.

∴ ∠BCD = $1/2$ × ∠BOD

= $1/2$ × 74° = 37°

Answer: (a)

Let ED = x

and area of rectangle ABCD

= AB × BC = 8 × 6 = 48 units

Now, AC = $√{8^2 + 6^2}$ = 10

In ΔAED, $AE^2 + ED^2 = AD^2$

$AE^2 = AD^2 – x^2 = 36 – x^2$ ...(i)

and in ΔCFD,

$CF^2 + DF^2 = CD^2$

⇒ $CF^2 = (8)^2 – (10 – x)^2$ ... (ii)

From eqs. (i) and (ii), we get

36 – $x^2 = 64 – (10 – x)^2$ (∵ AE = FC)

⇒ 36 – $x^2 = 64 – (100 + x^2$ – 20x)

(because AECE is rectangle)

⇒ 20x= 72 ⇒ x = ${18}/5$

From eq. (i) $AE^2 = 36 - ({18}/5)^2$

$AE^2 = 36 - {324}/{25} = {900 - 324}/{25}$

⇒ $AE^2 = {576}/{25}$

⇒ AE = ${24}/5$

∴ ${\text"Area of rectangle ABCD"}/{\text"Area of rectangle AEFC"} = ∼ {8 × 6}/{10 × {24}/5}$ = 1

Answer: (d)

Since, secants ∠A and BN are intersecting at an exterior point M, then

LM × AM = BM × NM

⇒ ${MA}/{MB} = {MN}/{LM}$ < 1

Answer: (c)

∠ADB= ∠ACB = 60° (angles in the same segment are equal)

Answer: (d)

Perimeter P = 10 cm

Area A = 4 $cm^2$

2 (l + B) = 10

B = 5 – l

l × B = 4 $cm^2$

l (5 – l) = 4

5l – $l^2$ = 4

$l^2$ – 5l + 4 = 0

$l^2$ – l – 4l + 4 = 0

(l – 1) (l – 4) = 0

l = 4, l ≠ 1

B = 5 – 4 = 1