Title 3. Concept of similarity.pdf ✅

CONCEPT OF SIMILARITY TRIANGLE 06 Class 10th Mathematics (Answers Key) 1 Both A and B. 2 PQR similarity criterion 3 AAA similarity criterion 4 Yes, by AAA similarity criterion. 5 They are not the mirror image of one another. 6 From the given triangles, ABC and PQR, we can get AB RQ = 3.8 7.6 = 1 2 BC QP = 6 12 = 1 2 CA PR = 3√3 6√3 = 1 2 Therefore, AB RQ = BC QP = CA PR Hence, by using the SSS similarity criterion for a triangle, we can write ∆ABC ~ ∆RQP (i.e.) ∆ABC is similar to ∆RQP. By using corresponding angles of similar triangles, ∠C = ∠P ∠C = 180° - ∠A - ∠B (Using the angle sum property of triangle). ∠C= 180° - 80° - 60° ∠C= 40°. Since, ∠C = ∠P, the value of ∠p is 40°. 7 In ΔABC and ΔPQR ∠A = ∠Q = 85° ∠B = ∠P = 60° and ∠C = ∠R = 35° So, ΔPQR ∼ ΔLMN [by AAA similarity] 8 ABCD is a trapezium with AB∥CD and diagonals AB and CD intersecting at O. ⇒ In △OAB and △OCD ⇒ ∠AOB = ∠DOC [ Vertically opposite angles] ⇒ ∠ABO = ∠CDO [ Alternate angles] ⇒ ∠BAO = ∠OCD [ Alternate angles] ∴ △OAB ∼ △OCD [ AAA similarity] We know that if triangles are similar, their corresponding sides are in proportion. Page 1 Email: Phone No: Website: [email protected] +91 93299 65211 www.brisklearning.com Address: Office No. 605, CTS No. 658, Gayatri Bhuvan, A & B, Kandivali, Dattani Park, Thakur Village, Kandivali East, Mumbai, Maharashtra 400101
CONCEPT OF SIMILARITY TRIANGLE 06 Class 10th Mathematics ⇒ OA OC = OB OD 9 Given, ABCD is a trapezium, and side AB is parallel to DC .i.e. AB || DC. Diagonals AC and BD intersect each other at point O. From the point O, draw a line EO touching AD at E, such that, EO || DC || AB In ΔADC, OE || DC From the Basic Proportionality Theorem, AE ED = AO CO ....(i) Now, In ΔABD, OE || AB From the Basic Proportionality Theorem, DE EA = DO BO ...(ii) From equations (i) and (ii), AO CO = BO DO ⇒ AO BO = CO DO 10 Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR i.e. AB PQ = BC QR = AD PM We have to prove: ΔABC ~ ΔPQR As we know here, AB PQ = BC QR = AD PM AB PQ = 1 2 BC 1 2 QR = AD PM ... ... (i) Page 2 Email: Phone No: Website: [email protected] +91 93299 65211 www.brisklearning.com Address: Office No. 605, CTS No. 658, Gayatri Bhuvan, A & B, Kandivali, Dattani Park, Thakur Village, Kandivali East, Mumbai, Maharashtra 400101
CONCEPT OF SIMILARITY TRIANGLE 06 Class 10th Mathematics ⇒ AB PQ = BC QR = AD PM (D is the midpoint of BC. M is the midpoint of QR) ⇒ ΔABD ~ ΔPQM [SSS similarity criterion] ∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal] ⇒ ∠ABC = ∠PQR In ΔABC and ΔPQR AB PQ = BC QR .......(i) ∠ABC = ∠PQR ...... (ii) From equation (i) and (ii), we get, ΔABC ~ ΔPQR [SAS similarity criterion] Page 3 Email: Phone No: Website: [email protected] +91 93299 65211 www.brisklearning.com Address: Office No. 605, CTS No. 658, Gayatri Bhuvan, A & B, Kandivali, Dattani Park, Thakur Village, Kandivali East, Mumbai, Maharashtra 400101