Tag: geometry

Quizzes (32) Questions (16)

Questions Related to geometry

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Find the side of the square whose diagonal is $16 \sqrt 2$ cm.

  1. $4$ cm
  2. $16$ cm
  3. $8$ cm
  4. $16\sqrt 2$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know that,

1) All angles of a square are congruent. i.e $90^o$
2) Diagonal of a square bisects each of its angles.
Therefore, the square gets divided into $2$ triangles of degrees $45^o-45^o-90^o$
$\therefore \sin 45^o = \cfrac {\text {side}}{\text {hyp}}$ 
$\therefore \cfrac {1}{\sqrt 2} = \cfrac {\text {side}}{16 \sqrt 2}$
$\therefore$ side of the square $= 16$ cm.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

The sides of triangle are in A.P. and the greatest angle exceeds the least by 90. The sides are in the ratio _____________.

  1. $1 : 2 : \sqrt { 2 }$
  2. $1 : \sqrt { 3 } : 2$
  3. $\sqrt { 7 } + 1 : \sqrt { 7 } : \sqrt { 7 } - 1$
  4. $\sqrt { 3 } + 1 : 1 : \sqrt { 3 } - 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the sides be a-d, a, a+d. Using the law of cosines and the condition that the largest angle exceeds the smallest by 90 degrees, one can derive the ratio of the sides as sqrt(7)+1 : sqrt(7) : sqrt(7)-1.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

If  H is orthocenter of triangle PQR then PH + QH + RH is 

  1. QR cot P + PR cot Q + PQ cot R

  2. (pq + QR + RP) (cot P + cot Q + copt R)
  3. $\dfrac{1}{2r}(cot P + cotQ + cot R)$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In any triangle, the distance from the orthocenter to the vertices is given by 2R cos A, 2R cos B, and 2R cos C. Summing these and relating them to the side lengths and cotangents leads to the identity PH + QH + RH = QR cot P + PR cot Q + PQ cot R.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

ABC is a triangle right angle at B. D is a point on AC such that $\angle ABD = 45^0$. If AC =$6$ and AD =$2$ , then AB is 

  1. $\dfrac{6}{\sqrt{5}}$
  2. ${3}{\sqrt{2}}$
  3. $\dfrac{12}{\sqrt{5}}$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the area of triangle ABC as the sum of areas of ABD and BCD, or using trigonometry in right triangles, we find AB = 6/sqrt(5).

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

consider a triangle PQR in which the relation $ QR^2+PR^2=5*PQ^2$ holds. let G be the point of intersection of the medians PM and QN . then angle QGM is always

  1. less then 45 degree

  2. obtuse

  3. a right angle

  4. acute and larger than 45 degree
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Apollonius theorem and the properties of medians, the condition QR^2 + PR^2 = 5PQ^2 implies specific geometric constraints on the triangle, leading to the angle QGM being less than 45 degrees.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

If in a $\Delta ABC,\sin A=\sin^{2} B$ and $2\cos^{2}A=3\cos^{2}B$, then the $\Delta ABC$ is 

  1. Right angled

  2. Obtuse angled

  3. Isosceles

  4. Equilateral
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given sin A = sin^2 B and 2 cos^2 A = 3 cos^2 B, substituting cos^2 A = 1 - sin^2 A = 1 - sin^4 B into the second equation allows solving for sin^2 B, which leads to A = 90 degrees.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Which of the following  can be the sides of a right-angled triangle?

  1. $0.5cm, 1.2 cm, 1.3cm$
  2. $2.4cm, 3.2 cm, 7.9cm$
  3. $5.0cm, 5.25 cm, 7.25cm$
  4. $1.6cm, 3.0 cm, 3.4cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A right-angled triangle must satisfy the Pythagorean theorem: a^2 + b^2 = c^2. For 0.5, 1.2, 1.3: 0.25 + 1.44 = 1.69, which is 1.3^2.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

Let $\Delta _1$ denotes the area of the triangle formed by the vertices $(a^3m^3 _1, am _1), (a^3m^3 _2am _2), (a^3m^3 _3, am _3)$ and $\Delta _2$ denotes the area of the triangle formed by the vertices $(2am _1m _2, a^2(m^2 _1+m^2 _2))$, $(2am _2m _3, a^2(m^2 _2+m^2 _3))$ and $(2am _3m _1, a^2(m^2 _3+m^2 _1))$. Then $\dfrac{\Delta _1}{\Delta _2}(a > 0)$ equals?

  1. $\dfrac{a}{2}$
  2. $2a$
  3. $\dfrac{a^3}{8}$
  4. $8a^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Calculating the area of the triangles using the determinant formula for coordinates and simplifying the ratio yields a/2.

Multiple choice maths geometry similarity and right angled triangle angle theorems for a right angled triangle apollonius's theorem

P, Q, R are the points of intersection of a line 1 with sides BC, CA, AB of a $\Delta$ ABC 
respectively, then $\dfrac{BP}{PC} \dfrac{CQ}{QA} \dfrac{AR}{RB}$

  1. 1

  2. 2

  3. -1

  4. -2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a direct application of Menelaus' Theorem, which states that for a line intersecting the sides of a triangle, the product of the ratios of the segments is 1.