Triangles

Concept: The shaded area = Area of equilateral triangle - Sum of areas of three circular sectors. Step 1: Area of the equilateral triangle ( ) Side cm. . . Step 2: Area of the three circular sectors ( ) Each angle of an equilateral triangle is . This is the angle for each sector. Radius of each sector cm. . Area of one sector = . Area of three identical sectors = . . (Alternatively, the three sectors form a semicircle of radius 4 cm). Step 3: Area of the shaded region Area (shaded) = Area (shaded) = . Step 4: Round to 2 decimal places Rounding to two decimal places gives . Option (1) is , which is the closest. The slight difference is likely due to rounding in the provided values of or , or the options themselves.

Concept: This problem involves similarity of triangles using the SAS (Side-Angle-Side) similarity criterion. If two triangles are similar, their corresponding angles are equal. Step 1: Analyze the given information for similarity We are given: % Option (A) (One pair of corresponding angles is equal). % Option (B) AB : ED = AC : EF. This can be rewritten as . This means the ratio of sides including the angle A in (AB, AC) is equal to the ratio of sides including the angle E in (ED, EF). The conditions and satisfy the SAS similarity criterion. The angle ( or ) is included between the sides whose ratios are given. So, . Note the order of vertices for similarity: A corresponds to E. Side AB (adjacent to A) corresponds to side ED (adjacent to E). Side AC (adjacent to A) corresponds to side EF (adjacent to E). Thus, the similarity is . Step 2: Corresponding angles in similar triangles Since , their corresponding angles are equal:
(Given as )
(This is what we need to find, or related to it)
Step 3: Use the given angle We are given . From the similarity , we have . So, . Step 4: Find using the sum of angles in The sum of angles in any triangle is . In : . We know and we just found . Substitute these values: Therefore, .

Concept: Congruence of triangles ( ) means that two triangles have exactly the same size and shape. The order of vertices in the congruence statement is crucial as it indicates corresponding parts. Step 1: Understanding the congruence statement We are given . This statement implies the following correspondence between vertices:
Vertex A in corresponds to Vertex A in .
Vertex B in corresponds to Vertex C in .
Vertex C in corresponds to Vertex B in . Step 2: Identifying corresponding sides Since corresponding parts of congruent triangles are equal (CPCTC), we can equate the lengths of corresponding sides:
Side AB (from ) corresponds to Side AC (from ). Therefore, .
Side BC (from ) corresponds to Side CB (from ). Therefore, (This is trivial, as it's the same side).
Side AC (from ) corresponds to Side AB (from ). Therefore, (This is the same as the first conclusion). Step 3: Conclusion about the triangle From the correspondence, we found that . A triangle in which at least two sides are equal in length is called an isosceles triangle. Since , is an isosceles triangle with these two sides being equal. Step 4: Analyzing the options
(1) AB=AC: This matches our finding from the congruence.
(2) AB=BC: This is not necessarily true from the given congruence.
(3) AC=BC: This is not necessarily true from the given congruence.
(4) None of these: Incorrect, as option (1) is correct. Therefore, if , then is isosceles with .

Concept: This question refers to a well-known geometrical theorem called Varignon's Theorem, or a specific case of it. Varignon's Theorem states that the figure formed by joining the midpoints of the sides of any quadrilateral is a parallelogram. We need to determine the specific type of parallelogram formed when the original quadrilateral is a rectangle. Step 1: Properties of a Rectangle A rectangle is a parallelogram with four right angles. Its diagonals are equal in length and bisect each other. Step 2: Consider a rectangle ABCD Let P, Q, R, and S be the midpoints of sides AB, BC, CD, and DA respectively. We need to determine the type of quadrilateral PQRS. Step 3: Using the Midpoint Theorem In , S is the midpoint of AD and P is the midpoint of AB. By the Midpoint Theorem, SP is parallel to DB and . Similarly, in , R is the midpoint of CD and Q is the midpoint of BC. By the Midpoint Theorem, RQ is parallel to DB and . Thus, SP is parallel to RQ and . Also, in , P is the midpoint of AB and Q is the midpoint of BC. By the Midpoint Theorem, PQ is parallel to AC and . Similarly, in , S is the midpoint of AD and R is the midpoint of CD. By the Midpoint Theorem, SR is parallel to AC and . Thus, PQ is parallel to SR and . Since opposite sides are equal and parallel (SP=RQ and PQ=SR), PQRS is a parallelogram. Step 4: Properties of the sides of parallelogram PQRS We have and . In a rectangle, the diagonals are equal, i.e., . Therefore, . Since all four sides of the parallelogram PQRS are equal, PQRS is a rhombus. Step 5: When would it be a square? A rhombus becomes a square if its angles are . The angles of the rhombus PQRS would be if the diagonals of the original rectangle (AC and DB) are perpendicular. However, the diagonals of a general rectangle are not necessarily perpendicular (they are only perpendicular if the rectangle is a square). So, for a general rectangle, the figure formed is a rhombus. If the original rectangle is a square, then the figure formed by joining midpoints is also a square (which is a special type of rhombus). But the general case for a rectangle is a rhombus.

Concept: This problem uses the property of triangles on the same base and between the same parallel lines, and how areas of triangles are related when diagonals of a trapezium intersect. Step 1: Property of triangles on the same base and between the same parallels Triangles on the same base (or equal bases) and between the same parallel lines are equal in area. In trapezium ABCD, AB DC. Consider triangles and . They share the same base DC and are between the same parallel lines AB and DC. Therefore, Area( ) = Area( ). Step 2: Decompose the areas We can write the areas as: Area( ) = Area( ) + Area( ) Area( ) = Area( ) + Area( ) Step 3: Equate the areas and simplify Since Area( ) = Area( ), we have: Area( ) + Area( ) = Area( ) + Area( ) Subtract Area( ) from both sides of the equation: Area( ) = Area( ) Step 4: Conclusion The triangle which is equal in area to is . Visualization: Imagine the trapezium. The triangles and are also on the same base AB and between parallels AB and DC, so Area( ) = Area( ). Area( ) = Area( ) + Area( ) Area( ) = Area( ) + Area( ) Equating these: Area( ) + Area( ) = Area( ) + Area( ). Subtracting Area( ) from both sides gives Area( ) = Area( ).