Class 10 Exam  >  Class 10 Tests  >  Olympiad Preparation for Class 10  >  Math Olympiad Test: Triangles - 3 - Class 10 MCQ

Math Olympiad Test: Triangles - 3 - Class 10 MCQ


Test Description

10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Triangles - 3

Math Olympiad Test: Triangles - 3 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Triangles - 3 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Triangles - 3 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Triangles - 3 below.
Solutions of Math Olympiad Test: Triangles - 3 questions in English are available as part of our Olympiad Preparation for Class 10 for Class 10 & Math Olympiad Test: Triangles - 3 solutions in Hindi for Olympiad Preparation for Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt Math Olympiad Test: Triangles - 3 | 10 questions in 10 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study Olympiad Preparation for Class 10 for Class 10 Exam | Download free PDF with solutions
Math Olympiad Test: Triangles - 3 - Question 1

In rhombus ABCD, AB2 + BC2 + CD2 + DA2 =

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 1

Let the length of sides of rhombus be x, length of OC be x1 and length of OD be y1
Then, AB2 + BC2 + CD2 + DA2 = 4x2
Since, AC and DB bisect each other at O.
∴ AC = 2x1 and BD = 2y1
In ΔAOD, ΔDOC, ΔAOB, ΔBOC
4[x21 + y21] = 4x2
⇒ AC2 + BD2 = AB2 + BC2 + CD2 + AD2

Math Olympiad Test: Triangles - 3 - Question 2

In the given figure, DABC ~ DDCB, then AB × DB =

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 2

Since Δ’s ABC and DCB are similar,
∴ AB/AC = DC/DB ⇒ AB × DB = DC × AC

1 Crore+ students have signed up on EduRev. Have you? Download the App
Math Olympiad Test: Triangles - 3 - Question 3

P and Q are points on sides AB and AC respectively of ΔABC. If AP = 3 cm, PB = 6 cm, AQ = 5 cm and QC = 10 cm, then BC =

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 3

Since, 
∴ PQ || BC, [By converse of Thale’s theorem]
⇒ ∠APQ = ∠ABC and ∠AQP = ∠ACB (Corresponding angles)
∴ ΔAPQ ~ ΔABC [By AA similarity]

Math Olympiad Test: Triangles - 3 - Question 4

In the given figure, ∠BAC = ∠ADC, then CA/CB is

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 4

In Δ’s ABC and DAC,
∠BAC = ∠ADC (Given)
∠ACB = ∠ACD (Common angle)
∴ ΔABC ~ ΔDAC (By AA similarity)
⇒ AC/BC = DC/AC

Math Olympiad Test: Triangles - 3 - Question 5

ABC is right triangle, right angled at C. If p is the length of the perpendicular from C to AB and a, b, c have the usual meaning, then 

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 5

Area of ΔABC = 1/2 × c × p = 1/2 × b × a

But, c2 = a2 + b2 [By Pythagoras theorem]
∴ 

Math Olympiad Test: Triangles - 3 - Question 6

In the given figure, AD ⊥ BC, BE ⊥ AC, CF ⊥ AB, then AF2 + BD2 + CE2 =

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 6

In ΔODB and ΔODC, using Pythagoras theorem,
OB2 = OD2 + BD2 and OC2 = OD2 + CD2
∴ OC2 = BD2 – CD2 ...(i)
Similarly, we have
OC2 – OA2 = CE2 – AE2 ...(ii)
And OA2 – OB2 = AF2 – BF2 ...(iii)
Adding (i), (ii) and (iii), we get
BD2 + CE2 + AF2 – CD2 – AE2 – BF2 = 0
⇒ BD2 + CE2 + AF2 = CD2 + AE2 + BF2

Math Olympiad Test: Triangles - 3 - Question 7

ΔABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ΔDEF ~ ΔABC and EF = 4 cm, then perimeter of ΔDEF is

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 7


Since, ΔDEF ~ ΔABC [Given]

∴ Perimeter of ΔDEF = DE + EF + FD
= 6 cm + 4 cm + 5 cm = 15 cm

Math Olympiad Test: Triangles - 3 - Question 8

In the given ΔABC, AD⊥BC and ∠A is right angled. Then AD2 =

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 8

In Δ's CDA and BAD,
we have
∠BAD = 90° – ∠ABD
∠DAC = 90° – ∠BAD
= 90° – (90° – ∠ABD) = ∠ABD
⇒ ∠DAC = ∠ABD

Now, in Δ's BDA and ADC, we have
∠ABD = ∠CAD
∠BDA = ∠ADC [Each 90°]
∴ ΔBDA ~ ΔADC [By AA similarity]
⇒ 

Math Olympiad Test: Triangles - 3 - Question 9

In the given trapezium ABCD, AB || CD and AB = 2CD. If area of ΔAOB = 84 cm2, then the area of ΔCOD is

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 9

In ΔAOB and ΔCOD, we have
∠AOB = ∠COD
[Vertically opposite angles]

∠OAB = ∠OCD [Alternate interior angles]
∴ ΔAOB ~ ΔCOD [By AA similarity]
 [∵ AB = 2CD]

Math Olympiad Test: Triangles - 3 - Question 10

In the given figure, ∠ ABC = 90° and BD ⊥ AC . If BD = 8 cm, AD = 4 cm, then CD =

Detailed Solution for Math Olympiad Test: Triangles - 3 - Question 10

In ΔDBA and ΔDCB, we have
∠BDA = ∠CDB [Each 90°]
and ∠DBA = ∠DCB [Each = 90° – ∠A]
∴ ΔDBA ~ ΔDCB [By AA similarity]

13 videos|44 docs|187 tests
Information about Math Olympiad Test: Triangles - 3 Page
In this test you can find the Exam questions for Math Olympiad Test: Triangles - 3 solved & explained in the simplest way possible. Besides giving Questions and answers for Math Olympiad Test: Triangles - 3, EduRev gives you an ample number of Online tests for practice

Top Courses for Class 10

13 videos|44 docs|187 tests
Download as PDF

Top Courses for Class 10