Tag: similarity

Quizzes (13) Questions (85)

Questions Related to similarity

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

Basic proportionality theorem  is also known as

  1. Basic theorem

  2. Thales Theorem

  3. Potential theorem

  4. Unknown
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Basic proportionality theorem is also known as Thales Theorem. Thales was a famous Greek mathematician who gave an important truth relating two equiangular triangles.
Therefore, C is the correct answer.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

ABC is a triangle with AB = $13$ cm, BC =$14$ cm and CA=$15$ cm. AD and BE are the altitudes from A to B to BC and AC respectively. H is the point of intersection of the AD and BE. Then the ratio of $\frac { HD }{ HB } =$ 

  1. $\dfrac { 3 }{ 5 } $
  2. $\dfrac { 12 }{ 13 } $
  3. $\dfrac { 4 }{ 5 } $
  4. $\dfrac { 5 }{ 9 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
According to the question,
Triangle $BEC$ and triangle $BDH$ are similar, because they have the same angles this means that the sides  of these two triangles are in the same ratio.

So,

$\dfrac{{HD}}{{BD}} = \dfrac{{CE}}{{BC}}$

Note,However that $\displaystyle \frac{{CE}}{{BC}}$=$cosC$, 

Hence$\displaystyle \frac{{HD}}{{BD}}$=$cosC$, so we proceed to find $cosC$ using the cosine rule,

${c^2} = {a^2} + {b^2} - 2ab\cos C$

${13^2} = {14^2} + {15^2} - 2(14)(15)cosC$

$\cos C = \dfrac{{{{13}^2} - {{14}^2} - {{15}^2}}}{{ - 2 \times 14 \times 15}} = \dfrac{3}{5}$

$so\, \, \dfrac{{HD}}{{HB}} = \dfrac{3}{5}$












Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

In a triangle ABC, D and E are the point on the line segment BC and AC respectively, such that 2 BD = DC and 3 AE = 2 EC. The lines AD and BE meet at P,the line CP and AB F, then :

  1. AP:PD = 2:1

  2. BP : PE =4:

  3. BP:PE =5:4

  4. CP:PF = 7:2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using Van Schooten's theorem or mass point geometry: BD:DC = 1:2 and AE:EC = 2:3. Assign masses: C=1, B=2, A=3. Then D is at 3, E is at 5. P is the intersection of AD and BE. AP:PD = (mass at D)/(mass at A) = (2+1)/3 = 1. This calculation suggests a ratio of 1:1, but standard application of Menelaus theorem or mass points confirms the ratio.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

Let  $ABC$  be a triangle and  $D$  and  $E$  be two points on side  $AB$  such that  $AD = BE$.  If  $D P | B C$  and  $E Q | A C,$ then $P Q | A C.$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By Thales theorem, if DP || BC, then AD/AB = AP/AC. If EQ || AC, then BE/AB = BQ/BC. Given AD=BE, the ratios imply the segments are related, but PQ || AC is not necessarily true.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

$ABCD$ is a rectangl $P$ and $Q$ are poits on $AB$ and $BC$ respectively such that the area of triangle $APD=5$ area of triangle $PBQ=4$ and area of triangle $QCD=3$, all area in square units. THen the area of the triangle $DPQ$ in square units is

  1. $12$
  2. $\dfrac {20}{3}$
  3. $2\sqrt {21}$
  4. $\sqrt {21}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the rectangle sides be x and y. AP=a, PB=x-a, BQ=b, QC=y-b. Areas are 0.5*a*y=5, 0.5*(x-a)b=4, 0.5(y-b)*x=3. Solving this system for the area of DPQ (Area_rect - sum of triangles) yields 12.

Multiple choice maths similarity relation between perimeters of similar shapes basic proportionality theorem and its converse basic proportionality theorem

The areas of two similar triangle are $18\ cm^{2}$ and $32\ cm^{2}$ respectively. What is the ratio of their corresponding sides?

  1. $3:4$
  2. $4:3$
  3. $9:16$
  4. $16:9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The ratio of areas of similar triangles is the square of the ratio of their corresponding sides. sqrt(18/32) = sqrt(9/16) = 3/4.