Triangles

Correct option: 1 — only ii)

Correct option: 1 — 9 cm

Correct option: 4 — (All of the similarity criteria can be used.)

Correct option: 1 — 19.5 cm

Correct option: 2 — 21 cm

Correct option: 2 — 62.5 square units

Correct option: 2 — ΔABC ~ ΔXYZ

Correct option: 4 — 25/49

Correct option: 4 — (A) is true but (R) is false.

• Assumes AB = BC as p and writes the equation for the area of the triangle as: ½ × p² = 18. Using the above equation, finds p as 6 m. [0.5]

• Uses the Pythagoras theorem to find the length of string lights required (length of AC) as: √(p² + p²) = √(6² + 6²) = 6√2 m [0.5]

• Writes that the given ratio of sides is not correct. [0.5]

• Gives a valid reason. For example, the corresponding sides of ΔPQR and ΔSRT are QR and RT respectively. Hence, the ratio of the corresponding sides is 1. [0.5]

• Uses the basic proportionality theorem to write: 2/5 = 12/LM [0.5]

• Solves the above equation to find the length of LM as 30 cm. [0.5]

• Writes TU/TW = TS/TV. [0.5]

• Uses the converse of basic proportionality theorem to write VW || SU.

Writes that ∠TUS = ∠TWV as they are corresponding angles in parallel lines. (Award full marks if proved using similarity.) [0.5]

• Writes that for ΔPQR and ΔSQT:

i) ∠PQR = ∠SQT (common)

ii) PQ/SQ = QR/QT (using basic proportionality theorem) [1]

• Hence, concludes that ΔPQR ~ ΔSQT by SAS similarity criterion. [1]

• Writes that in ΔXYZ and ΔXUV,

i) ∠YXZ = ∠UXV (common)

ii) ∠YZX = ∠UVX = 30° (given)

Hence, concludes that ΔXYZ ~ ΔXUV by AA similarity criterion. [1]

• Uses similarity of triangles to write the relation of sides as: XY/XU = YZ/UV

Hence, concludes that p = qr/(r + s). [1]

• Writes any two for ΔDHE and ΔFHG:

i) ∠DHE = ∠FHG (Vertically opposite angles are equal.)

ii) ∠HDE = ∠HFG (Alternate interior angles are equal.)

iii) ∠HED = ∠HGF (Alternate interior angles are equal.) [1.5]

• Writes that ΔDHE and ΔFHG are similar using AAA similarity criterion. (Award full marks if AA similarity criterion is correctly used.) [0.5]

• Assumes the length of each side of the square to be p units, where p is a real number.

Uses Pythagoras's theorem to find the length of the hypotenuse as: √(p² + p²) = p√2 units.

Writes that the length of the hypotenuse does not equal to p. [1.5]

• Concludes that the triangle is not an equilateral right-angled triangle and his claim is incorrect. [0.5]

• Mentions that the bottle and its shadow and Sarthak and his shadow form similar triangles. [0.5]

• Identifies the corresponding sides of similar triangles and writes:

Height of water bottle / Length of the shadow of water bottle = Height of Sarthak / Length of the shadow of Sarthak

⇒ 24/30 = 150 / Length of the shadow of Sarthak [1]

• Solves the equation in Step 2 to find the length of Sarthak's shadow as 375/2 cm or 187.5 cm. [0.5]

• Writes that in ΔJKL and ΔMKN,

i) ∠LJK = ∠NMK (given)

ii) ∠JKL = ∠MKN (tangents to a circle are perpendicular at the point of contact)

Hence, concludes that ΔJKL ~ ΔMKN by AA similarity criterion. [1]

• Finds the ratio of the corresponding sides of ΔJKL and ΔMKN as KL/KN = 2/1 as KN is the radius and KL is the diameter. [0.5]

• Uses the ratio of corresponding sides of similar triangles writes JL/MN = 2/1 to get MN as 7.5 cm. [0.5]

• i) Uses basic proportionality theorem to write:

QX/QZ = RY/RZ

=> 10/8 = (b+1)/b [1]

• Solves the above equation to find the value of b as 4 cm. [0.5]

• Uses the value of b and finds the length of YZ as 9 cm. [0.5]

• ii) Uses basic proportionality theorem to write PX/PY = QX/QZ. [0.5]

• Uses steps 1 and 4 to show that PX/PY = RY/RZ. [0.5]

• i) Mentions ZQ/QY = ZP/PX and finds PQ || XY using converse of basic proportionality theorem. (Award full marks if another appropriate method is correctly used.) [0.5]

• Writes that in ΔPQZ and ΔXYZ,

♦ ∠PQZ = ∠XYZ = 90° (corresponding angles as PQ || XY)

♦ ∠PZQ = ∠XZY (common)

Hence, ΔPQZ ~ ΔXYZ using AA similarity criterion. [1]

• ii) Gives proof for either similarity or congruency of ΔPQY and ΔPQZ. For Example,

♦ PQ/PQ = QY/QZ = 1

♦ ∠PQY = ∠PQZ = 90°

Hence, ΔPQY ~ ΔPQZ using SAS similarity criterion. [1]

• Finds ∠PYQ = ∠PZQ = 30° as ΔPQY is similar to ΔPQZ. [0.5]

• i) Writes that she can know the width by using the properties of similar triangles. [0.5]

• ii) Proves that ΔEDC and ΔABC are similar. For example,

i) ∠DCE = ∠ACB

ii) CE/AC = CD/BC

Hence, using SAS similarity criterion, ΔEDC and ΔABC are similar. [1.5]

• Uses the above step to get the following equation,

CE/AC = CD/BC = DE/AB = 1/4

Solves it to find the width, AB = 6.8 cm. [1]

• Writes that in ΔUXY and ΔUVW:

i) UX/UV = UY/UW = 2/3 (given)

ii) ∠XUY = ∠VUW (common angle)

Hence, by SAS similarity criterion, ΔUXY and ΔUVW are similar. [1.5]

• Uses the ratio of the corresponding sides of similar triangles to write:

UX/UV = XY/VW

=> 2/3 = 7/VW [1]

• Solves the above equation to find the length of VW as 10.5 units. [0.5]

• i) Writes that for ΔDGF and ΔDHE,

♦ ∠DGF = ∠DHE = 90°

♦ ∠FDG = ∠EDH (Common)

Uses AA similarity criterion to prove that ΔDGF ~ ΔDHE. [1.5]

• ii) Writes that for ΔIHF and ΔIGE,

♦ ∠IHF = ∠IGE = 90°

♦ ∠HFI = ∠GEI (Corresponding angles of similar triangles, ΔDGF and ΔDHE)

Uses AA similarity criterion to prove ΔIHF ~ ΔIGE. [1.5]

• i) Mentions that Shahnawaz is correct. [0.5]

• Mentions that as per the SSS similarity criterion, the ratio of corresponding sides must be the same, which is true in this case. [0.5]

• Mentions that Dhruv is incorrect. [0.5]

• Mentions that as per the AAA similarity criterion, the corresponding angles must be equal. In ΔPQR and ΔMNO, ∠P and ∠O are not corresponding angles. Hence, AAA similarity criterion cannot be used. [1]

• ii) Mentions that Abhiniti is correct. [0.5]

• Uses the information from part i), ΔPQR is similar to ΔMNO to write:

PQ = 5/7 MN, QR = 5/7 NO and RP = 5/7 OM. [1]

• Writes the following,

Perimeter of ΔPQR / Perimeter of ΔMNO = (PQ + QR + RP) / (MN + NO + OM)

Simplifies the expression to find the ratio of the perimeter of ΔPQR and ΔMNO as 5/7. [1]

• Uses Pythagoras theorem in ΔPQR to find the length of PR as:

PR² = 5² + 12²

=> PR = 13 [1]

• Writes in ΔVUR and ΔPQR:

♦ ∠VUR = ∠PQR (Right angle)

♦ ∠VRU = ∠PRQ (Common angle)

Hence, by AA similarity criterion, ΔVUR ~ ΔPQR. [1]

• Writes that in ΔPTS and ΔPQR:

♦ ∠PTS = ∠PQR (Right angle)

♦ ∠TPS = ∠RPQ (Common angle)

Hence, by AA similarity criterion, ΔPTS ~ ΔPQR. [1]

• Uses properties of similar triangles to write:

i) UR/QR = VR/PR

ii) PT/PQ = PS/PR

Evaluates equation i) to find UR = 60/13 cm and equation ii) to find PT = 20/13 cm. [1.5]

• Finds the length of UT as 13 − 60/13 − 20/13 = 89/13 cm or 6 11/13 cm. [0.5]

• Writes the following for ΔSLP, ΔPMQ, ΔQNR and ΔROS:

∠SLP = ∠PMQ = ∠QNR = ∠ROS (Right angles)

All the non-hypotenuse sides of the triangles are equal. (P, Q, R and S are midpoints of the sides of a square.) [0.5]

• Writes that ΔSLP ~ ΔPMQ ~ ΔQNR ~ ΔROS by SAS congruency criterion.

Hence, concludes that all the triangles are similar triangles as they are congruent. (Award full marks if proved using suitable alternative method.) [0.5]