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

Answer: (b)

A line perpendicular to the given line, passing through the given point is the required locus.

Answer: (a)

Let sides of the square be a.

Then, AC = a $√2$ and AO = OC = $a/√2$

Here, AM = $a/2$

∴ LM = $a/√2 - a/2$ and OM = $a/2$

In ΔOML,

tan $θ/2 = {a/√2 - a/2}/{a/2} = {{√2 - 1}/2}/{1/2} = √2 - 1$

∴ ∼ tan θ = ${2 tan θ/2}/{1 - tan^2 θ/2} = {2(√2 - 1)}/{1 - (2 + 1 - 2 √2)}$

= ${2(√2 - 1)}/{1 - 3 + 2 √2} = {2(√2 - 1)}/{2√2 - 2}$

⇒ tan θ = 1

Answer: (a)

1. Perimeter of triangle is greater than the sum of 3 medians

Let ABC be the triangle and D, E and F are midpoints of BC, CA and AB respectively.

Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side, (Theorem)

Hence in ΔABD, AD is a median

⇒ AB + AC > 2(AD)

Similarly, we get

BC + AC > 2CF

BC + AB > 2BE

On adding the above inequations, we get

(AB + AC) + (BC + AC) + (BC + AB) > 2AD + 2CD + BE

2(AB + BC + AC) > 2(AD + BE + CF)

∴ AB + BC + AC > AD + BE + CF

2. In triangle ABD, AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the third side] ...(1)

Similarly,

In triangle ADC, AC + DC > AD ...(2)

Adding 1 and 2 we get,

AB + BD + AC + DC > AD + AD

⇒ AB + (BD + DC) + AC > 2AD

⇒ AB + BC + AC > 2AD

Hence, both statements 1 and 2 are true.

Answer: (c)

Given two chords AB and CD intersects at a point P inside the circle.

According to condition of circle if two chords intersects at P then

AP × PB = CP × CD

${AP}/{CP} = {PD}/{PB}$ ...(1)

Given two right angled triangle A'P'B' and C'Q'D' then

A'P' = AP, B'P' = BP, C'Q' = CP, D'Q' = DP

From (1) we get

${AP}/{CP} = {PD}/{PB} ⇒ {A'P'}/{C'Q'} = {D'Q'}/{B'P'}$

⇒ ΔA'P'B' ∼ ΔC'Q'D'

∵ their corresponding ratios are equal. Also we know that in similar triangles.

${\text"Area of ΔA'P'B'"}/{\text"Area of ΔC'Q'D'"} = ({A'P'}/{C'Q'})^2 = ({AP}/{CP})^2$

⇒ ${1/2 × A'P' ∼ × P'B'}/{1/2 × C'Q' × Q'D'} = ({AP}/{CP})^2 ⇒ {AP × BP}/{CP × DP} = ({AP}/{CP})^2$

We know that

${AP}/{CP} = {PD}/{DB}$

⇒ $({AP}/{CP}) = ({BP}/{DP}) = ({AP}/{CP})^2$

⇒ $({AP}/{CP}) = ({AP}/{CP}) =({AP}/{CP})^2$

⇒ $({AP}/{CP})^2 = ({AP}/{CP})^2$

⇒ Area of ΔA'P'B' = Area of ΔC'Q'D'

These triangles are not congruent because none of the criterion are satisfied by these triangles perimeter of triangle. Sum of all three sides. But in these triangles, all the sides are not equal, then their perimeter is not equal

⇒ Statement (1) and (3) is correct.

Answer: (c)

From a point on the circle only one tangent can be drawn to a circle.