Triangles | Chapter Assessment | English | Grade 10

Determine whether ∆ ABC ~ ∆DEF. If yes, by which criteria?

• Yes SSS criteria

• Yes AAA Criteria

• Yes SAS criteria

• No

Answer explanation

Let ∆ ABC and ∆ DEF be two similar triangles So, if ∆ ABC ~ ∆ DEF, then, AB / DE = BC / EF = AC / DF AB/DE = 12/9 = 4/3 BC/EF = 6 / 4.5 = 4/3 AC / DF = 6√3 /4.5√3 = 4/3 So, ∆ ABC ~ ∆ DEF, by SSS Criteria So the correct option is Option 1

Tags

CCSS.HSG.SRT.B.5

In the figure, if DE ‖ BC, find AD.

• 21.5 cm

• 2.4 cm

• 1.35 cm

• None of the above

Answer explanation

If DE ‖ BC, then by Basic Proportionality theorem, AD / DB = AE / EC AD/ 7.2 = 1.8 / 5.4 AD= 7.2 x 1.8/5.4 = 2.4 cm So the correct option is Option 2

Tags

CCSS.HSG.SRT.B.4

In the figure, if DE ‖ BC, which of the following statements are true? 1. ∆ AME ~ ∆ ANC 2. ∆ AMD ~ ∆ ANB 3. ∆ AME ~ ∆ AMD

• 1 and 2

• Only 1

• Only 3

• All

Answer explanation

If DE ‖ BC, then in ∆ AME and ∆ ANC ∠ AME = ∠ ANC ( Corresponding Angles) ∠ NAC is common Angle So, ∆ AME ~ ∆ ANC ( AA Criteria) Similarly , in ∆ AMD and ∆ ANB ∠ AMD = ∠ ANB ( Corresponding Angles) ∠ NAB is common Angle So, ∆ AMD ~ ∆ ANB ( AA Criteria) However, ∆ AME and ∆ AMD are not similar So the correct option is Option 1

Tags

CCSS.HSG.SRT.A.2

The lengths of the sides of a triangle are 16, 23, 31. If the perimeter of a similar triangle is 280, find the length of the longest side of that triangle.

• 92

• 7.75

• 124

• None of the above

Answer explanation

Let ∆ ABC and ∆ DEF be the two similar triangles In ∆ ABC , Perimeter = 16 + 23 +31 = 70 Largest side = 31 In ∆ DEF, Perimeter = 280 Let longest side be x So, if ∆ ABC ~ ∆ DEF, then, Perimeter of ∆ABC/Perimeter of ∆DEF = 31/x 70/280 = 31/x x = 31 x 280/ 70 = 124 So the correct option is Option 3

With reference to the figure below, find AC and CE.

• AC = 3 CE = 2

• AC = 2 CE = 3

• AC = 10 CE = 10

Answer explanation

In ∆ CBA and ∆CDE, ∠ DCE = ∠ BCA (Vertically opposite angles) ∠ CDE = ∠ CBA = 90⁰ So, ∆ CBA ~ ∆CDE, So AC/ CE = BC /CD 12 ─ x/ x ─ 7= 4/6 72 ─ 6x = 4x ─ 28 10x = 100 x = 10 So, AC = 12 ─ 10 =2 CE = 10 ─ 7 = 3 So the correct option is Option 2

Tags

CCSS.HSG.SRT.B.4

In the given figure, there are 2 isosceles triangles, which are similar. PQ is not parallel to BC. If PC = 4, AQ = 3, QB = 12, BC =15 and AP =PQ, find ar(∆AQP) / ar (∆ACB).

• 1/9

• 3/12

• 5/4

• 3/15

Answer explanation

In the given figure, ∆ AQP ~ ∆ ACB, So AQ/ AC = AP /AB 3/(4 + AP) = AP/3+12 AP ( AP + 4) =3 x 15 AP² + 4AP ─ 45 = 0 AP² + 9 AP ─ 5 AP ─ 45 = 0 (AP +9)(AP ─ 5) = 0 So AP = ─ 9, 5 Since length cannot be negative, so AP =5 Since, ∆ AQP ~ ∆ ACB, ar(∆AQP) / ar (∆ACB) = AP²/AB² So , ar(∆APQ) / ar (∆ABC) = 5²/15² = 1/9

Tags

CCSS.HSG.SRT.A.2